Find a general solution do examples. The order of the differential equation and its solutions, the Cauchy problem

First order differential equations. Solution examples.
Differential equations with separable variables

Differential Equations (DE). These two words usually terrify the average layman. Differential equations seem to be something outrageous and difficult to master for many students. Oooooo... differential equations How can I survive all this?

Such an opinion and such an attitude is fundamentally wrong, because in fact DIFFERENTIAL EQUATIONS ARE SIMPLE AND EVEN FUN. What do you need to know and be able to learn to solve differential equations? To successfully study diffures, you must be good at integrating and differentiating. The better the topics are studied Derivative of a function of one variable and Indefinite integral, the easier it will be to understand differential equations. I will say more, if you have more or less decent integration skills, then the topic is practically mastered! The more integrals various types you know how to decide - the better. Why? You have to integrate a lot. And differentiate. Also highly recommend learn to find.

In 95% of cases in control work there are 3 types of first-order differential equations: separable equations, which we will cover in this lesson; homogeneous equations and linear inhomogeneous equations. For beginners to study diffusers, I advise you to read the lessons in this sequence, and after studying the first two articles, it will not hurt to consolidate your skills in an additional workshop - equations that reduce to homogeneous.

There are even rarer types of differential equations: equations in total differentials, Bernoulli's equations, and some others. Of the last two types, the most important are equations in total differentials, because in addition to this DE, I am considering new material - partial integration.

If you only have a day or two left, then for ultra-fast preparation there is blitz course in pdf format.

So, the landmarks are set - let's go:

Let us first recall the usual algebraic equations. They contain variables and numbers. The simplest example: . What does it mean to solve an ordinary equation? This means to find set of numbers that satisfy this equation. It is easy to see that the children's equation has a single root: . For fun, let's do a check, substitute the found root into our equation:

- the correct equality is obtained, which means that the solution is found correctly.

Diffuses are arranged in much the same way!

Differential equation first order in general contains:
1) independent variable ;
2) dependent variable (function);
3) the first derivative of the function: .

In some equations of the 1st order, there may be no "x" or (and) "y", but this is not essential - important so that in DU was first derivative, and did not have derivatives of higher orders - , etc.

What means ? To solve a differential equation means to find set of all functions that satisfy this equation. Such a set of functions often has the form ( is an arbitrary constant), which is called general solution of the differential equation.

Example 1

Solve differential equation

Full ammunition. Where to begin solution?

First of all, you need to rewrite the derivative in a slightly different form. We recall the cumbersome notation, which many of you probably thought was ridiculous and unnecessary. It is it that rules in diffusers!

In the second step, let's see if it's possible split variables? What does it mean to separate variables? Roughly speaking, on the left side we need to leave only "games", a on the right side organize only x's. Separation of variables is carried out with the help of “school” manipulations: parentheses, transfer of terms from part to part with a sign change, transfer of factors from part to part according to the rule of proportion, etc.

Differentials and are full multipliers and active participants in hostilities. In this example, the variables are easily separated by flipping factors according to the rule of proportion:

Variables are separated. On the left side - only "Game", on the right side - only "X".

Next stage - differential equation integration. It's simple, we hang integrals on both parts:

Of course, integrals must be taken. In this case, they are tabular:

As we remember, a constant is assigned to any antiderivative. There are two integrals here, but it is enough to write the constant once (because a constant + a constant is still equal to another constant). In most cases, it is placed on the right side.

Strictly speaking, after the integrals are taken, the differential equation is considered to be solved. The only thing is that our “y” is not expressed through “x”, that is, the solution is presented in the implicit form. The implicit solution of a differential equation is called general integral of the differential equation. That is, is the general integral.

An answer in this form is quite acceptable, but is there a better option? Let's try to get common decision .

Please, remember the first technique, it is very common and often used in practical tasks: if a logarithm appears on the right side after integration, then in many cases (but by no means always!) it is also advisable to write the constant under the logarithm.

That is, INSTEAD OF records are usually written .

Why is this needed? And in order to make it easier to express "y". We use the property of logarithms . In this case:

Now logarithms and modules can be removed:

The function is presented explicitly. This is the general solution.

Answer: common decision: .

The answers to many differential equations are fairly easy to check. In our case, this is done quite simply, we take the found solution and differentiate it:

Then we substitute the derivative into the original equation:

- the correct equality is obtained, which means that the general solution satisfies the equation , which was required to be checked.

Giving a constant various meanings, you can get infinitely many private decisions differential equation. It is clear that any of the functions , , etc. satisfies the differential equation .

Sometimes the general solution is called family of functions. In this example, the general solution is a family of linear functions, or rather, a family of direct proportionalities.

After a detailed discussion of the first example, it is appropriate to answer a few naive questions about differential equations:

1)In this example, we managed to separate the variables. Is it always possible to do this? No not always. And even more often the variables cannot be separated. For example, in homogeneous first order equations must be replaced first. In other types of equations, for example, in a linear non-homogeneous equation of the first order, you need to use various tricks and methods to find a general solution. The separable variable equations that we are looking at in the first lesson are − simplest type differential equations.

2) Is it always possible to integrate a differential equation? No not always. It is very easy to come up with a "fancy" equation that cannot be integrated, in addition, there are integrals that cannot be taken. But such DEs can be solved approximately using special methods. D'Alembert and Cauchy guarantee... ...ugh, lurkmore.to I read a lot just now, I almost added "from the other world."

3) In this example, we have obtained a solution in the form of a general integral . Is it always possible to find a general solution from the general integral, that is, to express "y" in an explicit form? No not always. For example: . Well, how can I express "y" here ?! In such cases, the answer should be written as a general integral. In addition, sometimes a general solution can be found, but it is written so cumbersomely and clumsily that it is better to leave the answer in the form of a general integral

4) ...perhaps enough for now. In the first example, we met another important point , but in order not to cover the "dummies" with an avalanche of new information, I will leave it until the next lesson.

Let's not hurry. Another simple remote control and another typical solution:

Example 2

Find a particular solution of the differential equation that satisfies the initial condition

Solution: according to the condition it is required to find private solution DE that satisfies a given initial condition. This kind of questioning is also called Cauchy problem.

First, we find a general solution. There is no “x” variable in the equation, but this should not be embarrassing, the main thing is that it has the first derivative.

We rewrite the derivative in the required form:

Obviously, the variables can be divided, boys to the left, girls to the right:

We integrate the equation:

The general integral is obtained. Here I drew a constant with an accent star, the fact is that very soon it will turn into another constant.

Now we are trying to convert the general integral into a general solution (express "y" explicitly). We remember the old, good, school: . In this case:

The constant in the indicator looks somehow not kosher, so it is usually lowered from heaven to earth. In detail, it happens like this. Using the property of degrees, we rewrite the function as follows:

If is a constant, then is also some constant, redesignate it with the letter :

Remember the "demolition" of a constant is second technique, which is often used in the course of solving differential equations.

So the general solution is: Such a nice family of exponential functions.

At the final stage, you need to find a particular solution that satisfies the given initial condition . It's simple too.

What is the task? Need to pick up such the value of the constant to satisfy the condition .

You can arrange it in different ways, but the most understandable, perhaps, will be like this. In the general solution, instead of “x”, we substitute zero, and instead of “y”, two:



That is,

Standard design version:

Now we substitute the found value of the constant into the general solution:
– this is the particular solution we need.

Answer: private solution:

Let's do a check. Verification of a particular solution includes two stages:

First, it is necessary to check whether the found particular solution really satisfies the initial condition ? Instead of "x" we substitute zero and see what happens:
- yes, indeed, a deuce was obtained, which means that the initial condition is satisfied.

The second stage is already familiar. We take the resulting particular solution and find the derivative:

Substitute in the original equation:

- the correct equality is obtained.

Conclusion: the particular solution is found correctly.

Let's move on to more meaningful examples.

Example 3

Solve differential equation

Solution: We rewrite the derivative in the form we need:

Assessing whether variables can be separated? Can. We transfer the second term to the right side with a sign change:

And we flip the factors according to the rule of proportion:

The variables are separated, let's integrate both parts:

I must warn you, judgment day is coming. If you have not learned well indefinite integrals, solved few examples, then there is nowhere to go - you have to master them now.

The integral of the left side is easy to find, with the integral of the cotangent we deal with the standard technique that we considered in the lesson Integration of trigonometric functions In the past year:

On the right side, we have a logarithm, and, according to my first technical recommendation, the constant should also be written under the logarithm.

Now we try to simplify the general integral. Since we have only logarithms, it is quite possible (and necessary) to get rid of them. By using known properties maximally "pack" the logarithms. I will write in great detail:

The packaging is complete to be barbarously tattered:

Is it possible to express "y"? Can. Both parts must be squared.

But you don't have to.

Third technical advice: if to obtain a general solution you need to raise to a power or take roots, then In most cases you should refrain from these actions and leave the answer in the form of a general integral. The fact is that the general solution will look just awful - with big roots, signs and other trash.

Therefore, we write the answer as a general integral. It is considered good form to present it in the form, that is, on the right side, if possible, leave only a constant. It is not necessary to do this, but it is always beneficial to please the professor ;-)

Answer: general integral:

! Note: the general integral of any equation can be written in more than one way. Thus, if your result did not coincide with a previously known answer, then this does not mean that you solved the equation incorrectly.

The general integral is also checked quite easily, the main thing is to be able to find derivative of a function defined implicitly. Let's differentiate the answer:

We multiply both terms by:

And we divide by:

The original differential equation was obtained exactly, which means that the general integral was found correctly.

Example 4

Find a particular solution of the differential equation that satisfies the initial condition. Run a check.

This is an example for independent solution.

I remind you that the algorithm consists of two stages:
1) finding a general solution;
2) finding the required particular solution.

The check is also carried out in two steps (see the sample in Example No. 2), you need:
1) make sure that the particular solution found satisfies the initial condition;
2) check that a particular solution generally satisfies the differential equation.

Full solution and answer at the end of the lesson.

Example 5

Find a particular solution of a differential equation , satisfying the initial condition . Run a check.

Solution: First, let's find a general solution. This equation already contains ready-made differentials and , which means that the solution is simplified. Separating variables:

We integrate the equation:

The integral on the left is tabular, the integral on the right is taken the method of summing the function under the sign of the differential:

The general integral has been obtained, is it possible to successfully express the general solution? Can. We hang logarithms on both sides. Since they are positive, the modulo signs are redundant:

(I hope everyone understands the transformation, such things should already be known)

So the general solution is:

Let's find a particular solution corresponding to the given initial condition .
In the general solution, instead of “x”, we substitute zero, and instead of “y”, the logarithm of two:

More familiar design:

We substitute the found value of the constant into the general solution.

Answer: private solution:

Check: First, check if the initial condition is met:
- everything is good.

Now let's check whether the found particular solution satisfies the differential equation at all. We find the derivative:

Let's look at the original equation: – it is presented in differentials. There are two ways to check. It is possible to express the differential from the found derivative:

We substitute the found particular solution and the resulting differential into the original equation :

We use the basic logarithmic identity:

The correct equality is obtained, which means that the particular solution is found correctly.

The second way of checking is mirrored and more familiar: from the equation express the derivative, for this we divide all the pieces by:

And in the transformed DE we substitute the obtained particular solution and the found derivative . As a result of simplifications, the correct equality should also be obtained.

Example 6

Solve the differential equation. Express the answer as a general integral.

This is an example for self-solving, full solution and answer at the end of the lesson.

What difficulties await in solving differential equations with separable variables?

1) It is not always obvious (especially to a teapot) that variables can be separated. Consider a conditional example: . Here you need to take the factors out of brackets: and separate the roots:. How to proceed further is clear.

2) Difficulties in the integration itself. Integrals often arise not the simplest, and if there are flaws in the skills of finding indefinite integral, then it will be difficult with many diffusers. In addition, the compilers of collections and manuals are popular with the logic “since the differential equation is simple, then at least the integrals will be more complicated.”

3) Transformations with a constant. As everyone has noticed, a constant in differential equations can be handled quite freely, and some transformations are not always clear to a beginner. Let's look at another hypothetical example: . In it, it is advisable to multiply all the terms by 2: . The resulting constant is also some kind of constant, which can be denoted by: . Yes, and since there is a logarithm on the right side, it is advisable to rewrite the constant as another constant: .

The trouble is that they often do not bother with indices and use the same letter . As a result, the decision record takes the following form:

What heresy? Here are the errors! Strictly speaking, yes. However, from a substantive point of view, there are no errors, because as a result of the transformation of a variable constant, a variable constant is still obtained.

Or another example, suppose that in the course of solving the equation, a general integral is obtained. This answer looks ugly, so it is advisable to change the sign of each term: . Formally, there is again an error - on the right, it should be written . But it is informally implied that “minus ce” is still a constant ( which just as well takes on any values!), so putting a "minus" does not make sense and you can use the same letter.

I will try to avoid a careless approach, and still put down different indexes for constants when converting them.

Example 7

Solve the differential equation. Run a check.

Solution: This equation admits separation of variables. Separating variables:

We integrate:

The constant here does not have to be defined under the logarithm, since nothing good will come of it.

Answer: general integral:

Check: Differentiate the answer (implicit function):

We get rid of fractions, for this we multiply both terms by:

The original differential equation has been obtained, which means that the general integral has been found correctly.

Example 8

Find a particular solution of DE.
,

This is a do-it-yourself example. The only hint is that here you get a general integral, and, more correctly, you need to contrive to find not a particular solution, but private integral. Full solution and answer at the end of the lesson.

Ordinary differential equation called an equation that connects an independent variable, an unknown function of this variable and its derivatives (or differentials) of various orders.

The order of the differential equation is the order of the highest derivative contained in it.

In addition to ordinary ones, partial differential equations are also studied. These are equations relating independent variables, an unknown function of these variables and its partial derivatives with respect to the same variables. But we will only consider ordinary differential equations and therefore we will omit the word "ordinary" for brevity.

Examples of differential equations:

(1) ;

(3) ;

(4) ;

Equation (1) is of the fourth order, equation (2) is of the third order, equations (3) and (4) are of the second order, equation (5) is of the first order.

Differential equation n order does not have to explicitly contain a function, all its derivatives from first to n th order and an independent variable. It may not explicitly contain derivatives of some orders, a function, an independent variable.

For example, in equation (1) there are clearly no derivatives of the third and second orders, as well as functions; in equation (2) - second-order derivative and function; in equation (4) - independent variable; in equation (5) - functions. Only equation (3) explicitly contains all derivatives, the function, and the independent variable.

By solving the differential equation any function is called y = f(x), substituting which into the equation, it turns into an identity.

The process of finding a solution to a differential equation is called its integration.

Example 1 Find a solution to the differential equation.

Solution. We write this equation in the form . The solution is to find the function by its derivative. The original function, as is known from the integral calculus, is the antiderivative for, i.e.

That's what it is solution of the given differential equation . changing in it C, we will get different solutions. We found out that there are an infinite number of solutions to a first-order differential equation.

General solution of the differential equation n th order is its solution expressed explicitly with respect to the unknown function and containing n independent arbitrary constants, i.e.

The solution of the differential equation in example 1 is general.

Partial solution of the differential equation is called such a solution in which the arbitrary constants are given specific numerical values.

Example 2 Find the general solution of the differential equation and a particular solution for .

Solution. We integrate both parts of the equation such a number of times that the order of the differential equation is equal.

,

.

As a result, we got the general solution -

given third-order differential equation.

Now let's find a particular solution under the specified conditions. To do this, we substitute their values ​​instead of arbitrary coefficients and obtain

.

If, in addition to the differential equation, the initial condition is given in the form , then such a problem is called Cauchy problem . The values ​​and are substituted into the general solution of the equation and the value of an arbitrary constant is found C, and then a particular solution of the equation for the found value C. This is the solution to the Cauchy problem.

Example 3 Solve the Cauchy problem for the differential equation from Example 1 under the condition .

Solution. We substitute into the general solution the values ​​from the initial condition y = 3, x= 1. We get

We write down the solution of the Cauchy problem for the given differential equation of the first order:

Solving differential equations, even the simplest ones, requires good skills in integrating and taking derivatives, including complex functions. This can be seen in the following example.

Example 4 Find the general solution of the differential equation.

Solution. The equation is written in such a form that both sides can be integrated immediately.

.

We apply the method of integration by changing the variable (substitution). Let , then .

Required to take dx and now - attention - we do it according to the rules of differentiation of a complex function, since x and there is a complex function ("apple" - extract square root or, which is the same, raising to the power of "one second", and "minced meat" is the very expression under the root):

We find the integral:

Returning to the variable x, we get:

.

This is the general solution of this differential equation of the first degree.

Not only skills from the previous sections of higher mathematics will be required in solving differential equations, but also skills from elementary, that is, school mathematics. As already mentioned, in a differential equation of any order there may not be an independent variable, that is, a variable x. The knowledge about proportions that has not been forgotten (however, anyone has it like) from the school bench will help to solve this problem. This is the next example.

Recall the problem that we faced when finding definite integrals:

or dy = f(x)dx. Her solution:

and it boils down to calculating indefinite integral. In practice, a more difficult task is more common: to find a function y, if it is known that it satisfies a relation of the form

This relation relates the independent variable x, unknown function y and its derivatives up to the order n inclusive, are called .

A differential equation includes a function under the sign of derivatives (or differentials) of one order or another. The order of the highest is called the order (9.1) .

Differential Equations:

- first order

second order,

- fifth order, etc.

A function that satisfies a given differential equation is called its solution , or integral . To solve it means to find all its solutions. If for the desired function y succeeded in obtaining a formula that gives all solutions, then we say that we have found its general solution , or general integral .

Common decision contains n arbitrary constants and looks like

If a relation is obtained that relates x, y and n arbitrary constants, in a form not permitted with respect to y -

then such a relation is called the general integral of equation (9.1).

Cauchy problem

Each specific solution, i.e., each specific function that satisfies a given differential equation and does not depend on arbitrary constants, is called a particular solution , or private integral. To obtain particular solutions (integrals) from general ones, it is necessary to attach specific numerical values ​​to the constants.

The graph of a particular solution is called an integral curve. The general solution, which contains all particular solutions, is a family of integral curves. For a first-order equation, this family depends on one arbitrary constant; for the equation n th order - from n arbitrary constants.

The Cauchy problem is to find a particular solution to the equation n th order, satisfying n initial conditions:

which determine n constants с 1 , с 2 ,..., c n.

1st order differential equations

For an unresolved with respect to the derivative, the differential equation of the 1st order has the form

or for permitted relatively

Example 3.46. Find a general solution to the equation

Solution. Integrating, we get

where C is an arbitrary constant. If we give C specific numerical values, then we get particular solutions, for example,

Example 3.47. Consider an increasing amount of money deposited in the bank, subject to the accrual of 100 r compound interest per year. Let Yo be the initial amount of money, and Yx after the expiration x years. When interest is calculated once a year, we get

where x = 0, 1, 2, 3,.... When interest is calculated twice a year, we get

where x = 0, 1/2, 1, 3/2,.... When calculating interest n once a year and if x takes successively the values ​​0, 1/n, 2/n, 3/n,..., then

Denote 1/n = h , then the previous equality will look like:

With unlimited magnification n(at ) in the limit we come to the process of increasing the amount of money with continuous interest accrual:

Thus, it can be seen that with a continuous change x the law of change in the money supply is expressed by a differential equation of the 1st order. Where Y x is an unknown function, x- independent variable, r- constant. We solve this equation, for this we rewrite it as follows:

where , or , where P stands for e C .

From the initial conditions Y(0) = Yo , we find P: Yo = Pe o , whence, Yo = P. Therefore, the solution looks like:

Consider the second economic problem. Macroeconomic models are also described by linear differential equations of the 1st order, describing the change in income or output Y as a function of time.

Example 3.48. Let the national income Y increase at a rate proportional to its size:

and let, the deficit in government spending is directly proportional to income Y with a proportionality coefficient q. The deficit in spending leads to an increase in the national debt D:

Initial conditions Y = Yo and D = Do at t = 0. From the first equation Y= Yoe kt . Substituting Y we get dD/dt = qYoe kt . The general solution has the form
D = (q/ k) Yoe kt +С, where С = const, which is determined from the initial conditions. Substituting the initial conditions, we obtain Do = (q/k)Yo + C. So, finally,

D = Do +(q/k)Yo (e kt -1),

this shows that the national debt is increasing at the same relative rate k, which is the national income.

Consider the simplest differential equations n order, these are equations of the form

Its general solution can be obtained using n times of integration.

Example 3.49. Consider the example y """ = cos x.

Solution. Integrating, we find

The general solution has the form

Linear differential equations

In economics, they are of great use, consider the solution of such equations. If (9.1) has the form:

then it is called linear, where po(x), p1(x),..., pn(x), f(x) - predefined functions. If f(x) = 0, then (9.2) is called homogeneous, otherwise it is called non-homogeneous. The general solution of equation (9.2) is equal to the sum of any of its particular solutions y(x) and the general solution of the homogeneous equation corresponding to it:

If the coefficients p o (x), p 1 (x),..., p n (x) are constants, then (9.2)

(9.4) is called a linear differential equation with constant coefficients order n .

For (9.4) it has the form:

We can set without loss of generality p o = 1 and write (9.5) in the form

We will look for a solution (9.6) in the form y = e kx , where k is a constant. We have: ; y " = ke kx , y "" = k 2 e kx , ..., y (n) = kne kx . Substitute the obtained expressions into (9.6), we will have:

(9.7) is an algebraic equation, its unknown is k, it is called characteristic. The characteristic equation has degree n and n roots, among which there can be both multiple and complex. Let k 1 , k 2 ,..., k n be real and distinct, then are particular solutions (9.7), while the general

Consider a linear homogeneous differential equation of the second order with constant coefficients:

Its characteristic equation has the form

(9.9)

its discriminant D = p 2 - 4q, depending on the sign of D, three cases are possible.

1. If D>0, then the roots k 1 and k 2 (9.9) are real and different, and the general solution has the form:

Solution. Characteristic equation: k 2 + 9 = 0, whence k = ± 3i, a = 0, b = 3, the general solution is:

y = C 1 cos 3x + C 2 sin 3x.

Second-order linear differential equations are used to study a web-like economic model with stocks of goods, where the rate of change of price P depends on the size of the stock (see paragraph 10). If supply and demand are linear functions prices, that is

a - is a constant that determines the reaction rate, then the process of price change is described by a differential equation:

For a particular solution, you can take a constant

which has the meaning of the equilibrium price. Deviation satisfies the homogeneous equation

(9.10)

The characteristic equation will be the following:

In case, the term is positive. Denote . The roots of the characteristic equation k 1,2 = ± i w, so the general solution (9.10) has the form:

where C and arbitrary constants, they are determined from the initial conditions. We have obtained the law of price change in time:

I. Ordinary differential equations

1.1. Basic concepts and definitions

A differential equation is an equation that relates an independent variable x, the desired function y and its derivatives or differentials.

Symbolically, the differential equation is written as follows:

F(x,y,y")=0, F(x,y,y")=0, F(x,y,y",y",.., y(n))=0

A differential equation is called ordinary if the desired function depends on one independent variable.

By solving the differential equation is called such a function that turns this equation into an identity.

The order of the differential equation is the order of the highest derivative in this equation

Examples.

1. Consider the first order differential equation

The solution to this equation is the function y = 5 ln x. Indeed, by substituting y" into the equation, we get - an identity.

And this means that the function y = 5 ln x– is the solution of this differential equation.

2. Consider the second order differential equation y" - 5y" + 6y = 0. The function is the solution to this equation.

Really, .

Substituting these expressions into the equation, we get: , - identity.

And this means that the function is the solution of this differential equation.

Integration of differential equations is the process of finding solutions to differential equations.

General solution of the differential equation is called a function of the form , which includes as many independent arbitrary constants as the order of the equation.

Partial solution of the differential equation is called the solution obtained from the general solution for different numerical values ​​of arbitrary constants. The values ​​of arbitrary constants are found at certain initial values ​​of the argument and function.

The graph of a particular solution of a differential equation is called integral curve.

Examples

1. Find a particular solution to a first-order differential equation

xdx + ydy = 0, if y= 4 at x = 3.

Solution. Integrating both sides of the equation, we get

Comment. An arbitrary constant C obtained as a result of integration can be represented in any form convenient for further transformations. In this case, taking into account the canonical equation of the circle, it is convenient to represent an arbitrary constant С in the form .

is the general solution of the differential equation.

A particular solution of an equation that satisfies the initial conditions y = 4 at x = 3 is found from the general by substituting the initial conditions into the general solution: 3 2 + 4 2 = C 2 ; C=5.

Substituting C=5 into the general solution, we get x2+y2 = 5 2 .

This is a particular solution of the differential equation obtained from the general solution under given initial conditions.

2. Find the general solution of the differential equation

The solution of this equation is any function of the form , where C is an arbitrary constant. Indeed, substituting into the equations, we obtain: , .

Therefore, this differential equation has an infinite number of solutions, since for different values ​​of the constant C, the equality determines different solutions of the equation.

For example, by direct substitution, one can verify that the functions are solutions of the equation .

A problem in which it is required to find a particular solution to the equation y" = f(x, y) satisfying the initial condition y(x0) = y0, is called the Cauchy problem.

Equation solution y" = f(x, y), satisfying the initial condition, y(x0) = y0, is called a solution to the Cauchy problem.

The solution of the Cauchy problem has a simple geometric meaning. Indeed, according to these definitions, to solve the Cauchy problem y" = f(x, y) on condition y(x0) = y0, means to find the integral curve of the equation y" = f(x, y) which goes through given point M0 (x0,y 0).

II. First order differential equations

2.1. Basic concepts

A first-order differential equation is an equation of the form F(x,y,y") = 0.

The first order differential equation includes the first derivative and does not include higher order derivatives.

The equation y" = f(x, y) is called a first-order equation solved with respect to the derivative.

A general solution of a first-order differential equation is a function of the form , which contains one arbitrary constant.

Example. Consider a first order differential equation.

The solution to this equation is the function .

Indeed, replacing in this equation with its value, we obtain

that is 3x=3x

Therefore, the function is a general solution of the equation for any constant C.

Find a particular solution of this equation that satisfies the initial condition y(1)=1 Substituting initial conditions x=1, y=1 into the general solution of the equation , we obtain whence C=0.

Thus, we obtain a particular solution from the general one by substituting into this equation, the resulting value C=0 is a private decision.

2.2. Differential equations with separable variables

A differential equation with separable variables is an equation of the form: y"=f(x)g(y) or through differentials , where f(x) and g(y) are given functions.

For those y, for which , the equation y"=f(x)g(y) is equivalent to the equation in which the variable y is present only on the left side, and the variable x is present only on the right side. They say, "in the equation y"=f(x)g(y separating the variables.

Type equation is called a separated variable equation.

After integrating both parts of the equation on x, we get G(y) = F(x) + C is the general solution of the equation, where G(y) and F(x) are some antiderivatives, respectively, of functions and f(x), C arbitrary constant.

Algorithm for solving a first-order differential equation with separable variables

Example 1

solve the equation y" = xy

Solution. Derivative of a function y" replace with

we separate the variables

Let's integrate both parts of the equality:

Example 2

2yy" = 1- 3x 2, if y 0 = 3 at x0 = 1

This is a separated variable equation. Let's represent it in differentials. To do this, we rewrite this equation in the form From here

Integrating both parts of the last equality, we find

Substituting initial values x 0 = 1, y 0 = 3 find FROM 9=1-1+C, i.e. C = 9.

Therefore, the desired partial integral will be or

Example 3

Write an equation for a curve passing through a point M(2;-3) and having a tangent with a slope

Solution. According to the condition

This is a separable variable equation. Dividing the variables, we get:

Integrating both parts of the equation, we get:

Using the initial conditions, x=2 and y=-3 find C:

Therefore, the desired equation has the form

2.3. Linear differential equations of the first order

A first-order linear differential equation is an equation of the form y" = f(x)y + g(x)

where f(x) and g(x)- some given functions.

If a g(x)=0 then the linear differential equation is called homogeneous and has the form: y" = f(x)y

If then the equation y" = f(x)y + g(x) called heterogeneous.

General solution of a linear homogeneous differential equation y" = f(x)y given by the formula: where FROM is an arbitrary constant.

In particular, if C \u003d 0, then the solution is y=0 If the linear homogeneous equation has the form y" = ky where k is some constant, then its general solution has the form: .

General solution of a linear inhomogeneous differential equation y" = f(x)y + g(x) given by the formula ,

those. is equal to the sum of the general solution of the corresponding linear homogeneous equation and the particular solution of this equation.

For a linear inhomogeneous equation of the form y" = kx + b,

where k and b- some numbers and a particular solution will be a constant function . Therefore, the general solution has the form .

Example. solve the equation y" + 2y +3 = 0

Solution. We represent the equation in the form y" = -2y - 3 where k=-2, b=-3 The general solution is given by the formula .

Therefore, where C is an arbitrary constant.

2.4. Solution of linear differential equations of the first order by the Bernoulli method

Finding a General Solution to a First-Order Linear Differential Equation y" = f(x)y + g(x) reduces to solving two differential equations with separated variables using the substitution y=uv, where u and v- unknown functions from x. This solution method is called the Bernoulli method.

Algorithm for solving a first-order linear differential equation

y" = f(x)y + g(x)

1. Enter a substitution y=uv.

2. Differentiate this equality y"=u"v + uv"

3. Substitute y and y" into this equation: u"v + uv" =f(x)uv + g(x) or u"v + uv" + f(x)uv = g(x).

4. Group the terms of the equation so that u take it out of brackets:

5. From the bracket, equating it to zero, find the function

This is a separable equation:

Divide the variables and get:

Where . .

6. Substitute the received value v into the equation (from item 4):

and find the function This is a separable equation:

7. Write the general solution in the form: , i.e. .

Example 1

Find a particular solution to the equation y" = -2y +3 = 0 if y=1 at x=0

Solution. Let's solve it with substitution y=uv,.y"=u"v + uv"

Substituting y and y" into this equation, we get

Grouping the second and third terms on the left side of the equation, we take out the common factor u out of brackets

We equate the expression in brackets to zero and, having solved the resulting equation, we find the function v = v(x)

We got an equation with separated variables. We integrate both parts of this equation: Find the function v:

Substitute the resulting value v into the equation We get:

This is a separated variable equation. We integrate both parts of the equation: Let's find the function u = u(x,c) Let's find a general solution: Let us find a particular solution of the equation that satisfies the initial conditions y=1 at x=0:

III. Higher order differential equations

3.1. Basic concepts and definitions

A second-order differential equation is an equation containing derivatives not higher than the second order. In the general case, the second-order differential equation is written as: F(x,y,y",y") = 0

The general solution of a second-order differential equation is a function of the form , which includes two arbitrary constants C1 and C2.

A particular solution of a second-order differential equation is a solution obtained from the general one for some values ​​of arbitrary constants C1 and C2.

3.2. Linear homogeneous differential equations of the second order with constant ratios.

Linear homogeneous differential equation of the second order with constant coefficients is called an equation of the form y" + py" + qy = 0, where p and q are constant values.

Algorithm for solving second-order homogeneous differential equations with constant coefficients

1. Write the differential equation in the form: y" + py" + qy = 0.

2. Compose its characteristic equation, denoting y" through r2, y" through r, y in 1: r2 + pr +q = 0

A differential equation is an equation that includes a function and one or more of its derivatives. In most practical problems, functions are physical quantities, the derivatives correspond to the rates of change of these quantities, and the equation determines the relationship between them.

This article discusses methods for solving some types of ordinary differential equations, the solutions of which can be written in the form elementary functions, that is, polynomial, exponential, logarithmic and trigonometric functions, as well as their inverse functions. Many of these equations are found in real life, although most other differential equations cannot be solved by these methods, and for them the answer is written as special functions or power series, or found by numerical methods.

To understand this article, you need to know differential and integral calculus, as well as have some understanding of partial derivatives. It is also recommended to know the basics of linear algebra as applied to differential equations, especially second-order differential equations, although knowledge of differential and integral calculus is sufficient to solve them.

Preliminary information

• Differential equations have an extensive classification. This article talks about ordinary differential equations, that is, about equations that include a function of one variable and its derivatives. Ordinary differential equations are much easier to understand and solve than partial differential equations, which include functions of several variables. This article does not consider partial differential equations, since the methods for solving these equations are usually determined by their specific form.
  • Below are some examples of ordinary differential equations.
    • d y d x = k y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=ky)
    • d 2 x d t 2 + k x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+kx=0)
  • Below are some examples of partial differential equations.
    • ∂ 2 f ∂ x 2 + ∂ 2 f ∂ y 2 = 0 (\displaystyle (\frac (\partial ^(2)f)(\partial x^(2)))+(\frac (\partial ^(2 )f)(\partial y^(2)))=0)
    • ∂ u ∂ t − α ∂ 2 u ∂ x 2 = 0 (\displaystyle (\frac (\partial u)(\partial t))-\alpha (\frac (\partial ^(2)u)(\partial x ^(2)))=0)
• Order differential equation is determined by the order of the highest derivative included in this equation. The first of the above ordinary differential equations is of the first order, while the second is of the second order. Degree differential equation is called highest degree, to which one of the terms of this equation is raised.
  • For example, the equation below is third order and second power.
    • (d 3 y d x 3) 2 + d y d x = 0 (\displaystyle \left((\frac ((\mathrm (d) )^(3)y)((\mathrm (d) )x^(3)))\ right)^(2)+(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0)
• The differential equation is linear differential equation if the function and all its derivatives are in the first power. Otherwise, the equation is nonlinear differential equation. Linear differential equations are remarkable in that linear combinations can be made from their solutions, which will also be solutions to this equation.
  • Below are some examples of linear differential equations.
  • Below are some examples of non-linear differential equations. The first equation is non-linear due to the sine term.
    • d 2 θ d t 2 + g l sin ⁡ θ = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)\theta )((\mathrm (d) )t^(2)))+( \frac (g)(l))\sin \theta =0)
    • d 2 x d t 2 + (d x d t) 2 + t x 2 = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+ \left((\frac ((\mathrm (d) )x)((\mathrm (d) )t))\right)^(2)+tx^(2)=0)
• Common decision ordinary differential equation is not unique, it includes arbitrary constants of integration. In most cases, the number of arbitrary constants is equal to the order of the equation. In practice, the values ​​of these constants are determined by given initial conditions, that is, by the values ​​of the function and its derivatives at x = 0. (\displaystyle x=0.) The number of initial conditions that are needed to find private decision differential equation, in most cases is also equal to the order of this equation.
  • For example, this article will look at solving the equation below. This is a second order linear differential equation. Its general solution contains two arbitrary constants. To find these constants, it is necessary to know the initial conditions at x (0) (\displaystyle x(0)) and x′ (0) . (\displaystyle x"(0).) Usually the initial conditions are given at the point x = 0 , (\displaystyle x=0,), although this is not required. This article will also consider how to find particular solutions for given initial conditions.
    • d 2 x d t 2 + k 2 x = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)x)((\mathrm (d) )t^(2)))+k^(2 )x=0)
    • x (t) = c 1 cos ⁡ k x + c 2 sin ⁡ k x (\displaystyle x(t)=c_(1)\cos kx+c_(2)\sin kx)

Steps

Part 1

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1. Linear equations of the first order. This section discusses methods for solving linear differential equations of the first order in general and special cases, when some terms are equal to zero. Let's pretend that y = y (x) , (\displaystyle y=y(x),) p (x) (\displaystyle p(x)) and q (x) (\displaystyle q(x)) are functions x . (\displaystyle x.)

   D y d x + p (x) y = q (x) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+p(x)y=q(x ))
   

   P (x) = 0. (\displaystyle p(x)=0.) According to one of the main theorems of mathematical analysis, the integral of the derivative of a function is also a function. Thus, it is enough to simply integrate the equation to find its solution. In this case, it should be taken into account that when calculating the indefinite integral, an arbitrary constant appears.

   • y (x) = ∫ q (x) d x (\displaystyle y(x)=\int q(x)(\mathrm (d) )x)

   Q (x) = 0. (\displaystyle q(x)=0.) We use the method separation of variables. In this case, different variables are transferred to different sides of the equation. For example, you can transfer all members from y (\displaystyle y) into one, and all members with x (\displaystyle x) to the other side of the equation. Members can also be moved d x (\displaystyle (\mathrm (d) )x) and d y (\displaystyle (\mathrm (d) )y), which are included in the expressions of derivatives, however, it should be remembered that these are just symbol, which is convenient for differentiating complex function. A discussion of these terms, which are called differentials, is outside the scope of this article.

   • First, you need to move the variables on opposite sides of the equals sign.
     • 1 y d y = − p (x) d x (\displaystyle (\frac (1)(y))(\mathrm (d) )y=-p(x)(\mathrm (d) )x)
   • We integrate both sides of the equation. After integration, arbitrary constants appear on both sides, which can be transferred to the right side of the equation.
     • ln ⁡ y = ∫ − p (x) d x (\displaystyle \ln y=\int -p(x)(\mathrm (d) )x)
     • y (x) = e − ∫ p (x) d x (\displaystyle y(x)=e^(-\int p(x)(\mathrm (d) )x))
   • Example 1.1. In the last step, we used the rule e a + b = e a e b (\displaystyle e^(a+b)=e^(a)e^(b)) and replaced e C (\displaystyle e^(C)) on the C (\displaystyle C), because it is also an arbitrary constant of integration.
     • d y d x − 2 y sin ⁡ x = 0 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))-2y\sin x=0)
     • 1 2 y d y = sin ⁡ x d x 1 2 ln ⁡ y = - cos ⁡ x + C ln ⁡ y = - 2 cos ⁡ x + C y (x) = C e )(\frac (1)(2y))(\mathrm (d) )y&=\sin x(\mathrm (d) )x\\(\frac (1)(2))\ln y&=-\cos x+C\\\ln y&=-2\cos x+C\\y(x)&=Ce^(-2\cos x)\end(aligned)))

   P (x) ≠ 0 , q (x) ≠ 0. (\displaystyle p(x)\neq 0,\ q(x)\neq 0.) To find the general solution, we introduced integrating factor as a function of x (\displaystyle x) to reduce the left side to a common derivative and thus solve the equation.

   • Multiply both sides by μ (x) (\displaystyle \mu (x))
     • μ d y d x + μ p y = μ q (\displaystyle \mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py=\mu q)
   • To reduce the left side to a common derivative, the following transformations must be made:
     • d d x (μ y) = d μ d x y + μ d y d x = μ d y d x + μ p y (\displaystyle (\frac (\mathrm (d) )((\mathrm (d) )x))(\mu y)=(\ frac ((\mathrm (d) )\mu )((\mathrm (d) )x))y+\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x)) =\mu (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+\mu py)
   • The last equality means that d μ d x = μ p (\displaystyle (\frac ((\mathrm (d) )\mu )((\mathrm (d) )x))=\mu p). This is an integrating factor that is sufficient to solve any first order linear equation. Now we can derive a formula for solving this equation with respect to µ , (\displaystyle \mu ,) although for training it is useful to do all the intermediate calculations.
     • μ (x) = e ∫ p (x) d x (\displaystyle \mu (x)=e^(\int p(x)(\mathrm (d) )x))
   • Example 1.2. In this example, we consider how to find a particular solution to a differential equation with given initial conditions.
     • t d y d t + 2 y = t 2 , y (2) = 3 (\displaystyle t(\frac ((\mathrm (d) )y)((\mathrm (d) )t))+2y=t^(2) ,\quad y(2)=3)
     • d y d t + 2 t y = t (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )t))+(\frac (2)(t))y=t)
     • μ (x) = e ∫ p (t) d t = e 2 ln ⁡ t = t 2 (\displaystyle \mu (x)=e^(\int p(t)(\mathrm (d) )t)=e ^(2\ln t)=t^(2))
     • d d t (t 2 y) = t 3 t 2 y = 1 4 t 4 + C y (t) = 1 4 t 2 + C t 2 (\displaystyle (\begin(aligned)(\frac (\mathrm (d) )((\mathrm (d) )t))(t^(2)y)&=t^(3)\\t^(2)y&=(\frac (1)(4))t^(4 )+C\\y(t)&=(\frac (1)(4))t^(2)+(\frac (C)(t^(2)))\end(aligned)))
     • 3 = y (2) = 1 + C 4 , C = 8 (\displaystyle 3=y(2)=1+(\frac (C)(4)),\quad C=8)
     • y (t) = 1 4 t 2 + 8 t 2 (\displaystyle y(t)=(\frac (1)(4))t^(2)+(\frac (8)(t^(2)) ))
   
   Solving linear equations of the first order (recorded by Intuit - National Open University).

2. Nonlinear First Order Equations. In this section, methods for solving some nonlinear differential equations of the first order are considered. Although there is no general method for solving such equations, some of them can be solved using the methods below.

   D y d x = f (x , y) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=f(x,y))
   d y d x = h (x) g (y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=h(x)g(y).) If the function f (x , y) = h (x) g (y) (\displaystyle f(x,y)=h(x)g(y)) can be divided into functions of one variable, such an equation is called separable differential equation. In this case, you can use the above method:

   • ∫ d y h (y) = ∫ g (x) d x (\displaystyle \int (\frac ((\mathrm (d) )y)(h(y)))=\int g(x)(\mathrm (d) )x)
   • Example 1.3.
     • d y d x = x 3 y (1 + x 4) (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (x^(3))( y(1+x^(4)))))
     • ∫ y d y = ∫ x 3 1 + x 4 d x 1 2 y 2 = 1 4 ln ⁡ (1 + x 4) + C y (x) = 1 2 ln ⁡ (1 + x 4) + C (\displaystyle (\ begin(aligned)\int y(\mathrm (d) )y&=\int (\frac (x^(3))(1+x^(4)))(\mathrm (d) )x\\(\ frac (1)(2))y^(2)&=(\frac (1)(4))\ln(1+x^(4))+C\\y(x)&=(\frac ( 1)(2))\ln(1+x^(4))+C\end(aligned)))

   D y d x = g (x , y) h (x , y) . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (g(x,y))(h(x,y))).) Let's pretend that g (x , y) (\displaystyle g(x, y)) and h (x , y) (\displaystyle h(x, y)) are functions x (\displaystyle x) and y . (\displaystyle y.) Then homogeneous differential equation is an equation in which g (\displaystyle g) and h (\displaystyle h) are homogeneous functions the same degree. That is, the functions must satisfy the condition g (α x , α y) = α k g (x , y) , (\displaystyle g(\alpha x,\alpha y)=\alpha ^(k)g(x,y),) where k (\displaystyle k) is called the degree of homogeneity. Any homogeneous differential equation can be given by an appropriate change of variables (v = y / x (\displaystyle v=y/x) or v = x / y (\displaystyle v=x/y)) to convert to an equation with separable variables.

   • Example 1.4. The above description of homogeneity may seem obscure. Let's look at this concept with an example.
     • d y d x = y 3 − x 3 y 2 x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y^(3)-x^ (3))(y^(2)x)))
     • To begin with, it should be noted that this equation is non-linear with respect to y . (\displaystyle y.) We also see that in this case it is impossible to separate the variables. However, this differential equation is homogeneous, since both the numerator and denominator are homogeneous with a power of 3. Therefore, we can make a change of variables v=y/x. (\displaystyle v=y/x.)
     • d y d x = y x − x 2 y 2 = v − 1 v 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac (y)(x ))-(\frac (x^(2))(y^(2)))=v-(\frac (1)(v^(2))))
     • y = v x , d y d x = d v d x x + v (\displaystyle y=vx,\quad (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=(\frac ((\mathrm (d) )v)((\mathrm (d) )x))x+v)
     • d v d x x = − 1 v 2 . (\displaystyle (\frac ((\mathrm (d) )v)((\mathrm (d) )x))x=-(\frac (1)(v^(2))).) As a result, we have an equation for v (\displaystyle v) with shared variables.
     • v (x) = − 3 log ⁡ x + C 3 (\displaystyle v(x)=(\sqrt[(3)](-3\ln x+C)))
     • y (x) = x − 3 ln ⁡ x + C 3 (\displaystyle y(x)=x(\sqrt[(3)](-3\ln x+C)))

   D y d x = p (x) y + q (x) y n . (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=p(x)y+q(x)y^(n).) it Bernoulli differential equation- a special kind of nonlinear equation of the first degree, the solution of which can be written using elementary functions.

   • Multiply both sides of the equation by (1 − n) y − n (\displaystyle (1-n)y^(-n)):
     • (1 − n) y − n d y d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (1-n)y^(-n)(\frac ( (\mathrm (d) )y)((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))
   • We use the rule of differentiation of a complex function on the left side and transform the equation into linear equation relatively y 1 − n , (\displaystyle y^(1-n),) which can be solved by the above methods.
     • d y 1 − n d x = p (x) (1 − n) y 1 − n + (1 − n) q (x) (\displaystyle (\frac ((\mathrm (d) )y^(1-n)) ((\mathrm (d) )x))=p(x)(1-n)y^(1-n)+(1-n)q(x))

   M (x , y) + N (x , y) d y d x = 0. (\displaystyle M(x,y)+N(x,y)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))=0.) it total differential equation. It is necessary to find the so-called potential function φ (x , y) , (\displaystyle \varphi (x,y),), which satisfies the condition d φ d x = 0. (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=0.)

   • To fulfill this condition, it is necessary to have total derivative. The total derivative takes into account the dependence on other variables. To calculate the total derivative φ (\displaystyle \varphi ) on x , (\displaystyle x,) we assume that y (\displaystyle y) may also depend on x . (\displaystyle x.)
     • d φ d x = ∂ φ ∂ x + ∂ φ ∂ y d y d x (\displaystyle (\frac ((\mathrm (d) )\varphi )((\mathrm (d) )x))=(\frac (\partial \varphi )(\partial x))+(\frac (\partial \varphi )(\partial y))(\frac ((\mathrm (d) )y)((\mathrm (d) )x)))
   • Comparing terms gives us M (x , y) = ∂ φ ∂ x (\displaystyle M(x,y)=(\frac (\partial \varphi )(\partial x))) and N (x, y) = ∂ φ ∂ y . (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y)).) This is a typical result for equations with several variables, where the mixed derivatives of smooth functions are equal to each other. Sometimes this case is called Clairaut's theorem. In this case, the differential equation is an equation in total differentials if the following condition is satisfied:
     • ∂ M ∂ y = ∂ N ∂ x (\displaystyle (\frac (\partial M)(\partial y))=(\frac (\partial N)(\partial x)))
   • The method for solving equations in total differentials is similar to finding potential functions in the presence of several derivatives, which we will briefly discuss. First we integrate M (\displaystyle M) on x . (\displaystyle x.) Because the M (\displaystyle M) is a function and x (\displaystyle x), and y , (\displaystyle y,) when integrating, we get an incomplete function φ , (\displaystyle \varphi ,) labeled as φ ~ (\displaystyle (\tilde (\varphi ))). The result also includes the dependent on y (\displaystyle y) constant of integration.
     • φ (x , y) = ∫ M (x , y) d x = φ ~ (x , y) + c (y) (\displaystyle \varphi (x,y)=\int M(x,y)(\mathrm (d) )x=(\tilde (\varphi ))(x,y)+c(y))
   • After that, to get c (y) (\displaystyle c(y)) you can take the partial derivative of the resulting function with respect to y , (\displaystyle y,) equate the result N (x , y) (\displaystyle N(x, y)) and integrate. One can also integrate first N (\displaystyle N), and then take the partial derivative with respect to x (\displaystyle x), which will allow us to find an arbitrary function d(x). (\displaystyle d(x).) Both methods are suitable, and usually the simpler function is chosen for integration.
     • N (x , y) = ∂ φ ∂ y = ∂ φ ~ ∂ y + d c d y (\displaystyle N(x,y)=(\frac (\partial \varphi )(\partial y))=(\frac (\ partial (\tilde (\varphi )))(\partial y))+(\frac ((\mathrm (d) )c)((\mathrm (d) )y)))
   • Example 1.5. You can take partial derivatives and verify that the equation below is a total differential equation.
     • 3 x 2 + y 2 + 2 x y d y d x = 0 (\displaystyle 3x^(2)+y^(2)+2xy(\frac ((\mathrm (d) )y)((\mathrm (d) )x) )=0)
     • φ = ∫ (3 x 2 + y 2) d x = x 3 + x y 2 + c (y) ∂ φ ∂ y = N (x , y) = 2 x y + d c d y (\displaystyle (\begin(aligned)\varphi &=\int (3x^(2)+y^(2))(\mathrm (d) )x=x^(3)+xy^(2)+c(y)\\(\frac (\partial \varphi )(\partial y))&=N(x,y)=2xy+(\frac ((\mathrm (d) )c)((\mathrm (d) )y))\end(aligned)))
     • d c d y = 0 , c (y) = C (\displaystyle (\frac ((\mathrm (d) )c)((\mathrm (d) )y))=0,\quad c(y)=C)
     • x 3 + x y 2 = C (\displaystyle x^(3)+xy^(2)=C)
   • If the differential equation is not a total differential equation, in some cases you can find an integrating factor that will allow you to convert it to a total differential equation. However, such equations are rarely used in practice, and although the integrating factor exists, find it happens not easy, so these equations are not considered in this article.

Part 2

1. Homogeneous linear differential equations with constant coefficients. These equations are widely used in practice, so their solution is of paramount importance. In this case, we are not talking about homogeneous functions, but about the fact that there is 0 on the right side of the equation. In the next section, we will show how the corresponding heterogeneous differential equations. Below a (\displaystyle a) and b (\displaystyle b) are constants.

   D 2 y d x 2 + a d y d x + b y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. This differential equation is remarkable in that it can be solved very easily if you pay attention to what properties its solutions should have. It can be seen from the equation that y (\displaystyle y) and its derivatives are proportional to each other. From the previous examples, which were considered in the section on first-order equations, we know that only the exponential function has this property. Therefore, it is possible to put forward ansatz(an educated guess) about what the solution to the given equation will be.

   • The solution will take the form of an exponential function e r x , (\displaystyle e^(rx),) where r (\displaystyle r) is a constant whose value is to be found. Substitute this function into the equation and get the following expression
     • e r x (r 2 + a r + b) = 0 (\displaystyle e^(rx)(r^(2)+ar+b)=0)
   • This equation indicates that the product of an exponential function and a polynomial must be zero. It is known that the exponent cannot be equal to zero for any values ​​of the degree. Hence we conclude that the polynomial is equal to zero. Thus, we have reduced the problem of solving a differential equation to a much simpler problem of solving an algebraic equation, which is called the characteristic equation for a given differential equation.
     • r 2 + a r + b = 0 (\displaystyle r^(2)+ar+b=0)
     • r ± = − a ± a 2 − 4 b 2 (\displaystyle r_(\pm )=(\frac (-a\pm (\sqrt (a^(2)-4b)))(2)))
   • We have two roots. Since this differential equation is linear, its general solution is a linear combination of partial solutions. Since this is a second order equation, we know that this is really general solution, and there are no others. A more rigorous justification for this lies in the theorems on the existence and uniqueness of the solution, which can be found in textbooks.
   • A useful way to check if two solutions are linearly independent is to compute Wronskian. Wronskian W (\displaystyle W)- this is the determinant of the matrix, in the columns of which there are functions and their successive derivatives. The linear algebra theorem states that the functions in the Wronskian are linearly dependent if the Wronskian is equal to zero. In this section, we can test whether two solutions are linearly independent by making sure that the Wronskian is non-zero. The Wronskian is important in solving nonhomogeneous differential equations with constant coefficients by the parameter variation method.
     • w = | y 1 y 2 y 1 ′ y 2 ′ | (\displaystyle W=(\begin(vmatrix)y_(1)&y_(2)\\y_(1)"&y_(2)"\end(vmatrix)))
   • In terms of linear algebra, the set of all solutions of a given differential equation forms a vector space whose dimension is equal to the order of the differential equation. In this space, one can choose a basis from linearly independent decisions from each other. This is possible due to the fact that the function y (x) (\displaystyle y(x)) valid linear operator. Derivative is linear operator, since it transforms the space of differentiable functions into the space of all functions. Equations are called homogeneous in cases where for some linear operator L (\displaystyle L) it is required to find a solution to the equation L [ y ] = 0. (\displaystyle L[y]=0.)

   Let's now look at a few concrete examples. The case of multiple roots of the characteristic equation will be considered a little later, in the section on order reduction.

   If the roots r ± (\displaystyle r_(\pm )) are different real numbers, the differential equation has the following solution

   • y (x) = c 1 e r + x + c 2 e r − x (\displaystyle y(x)=c_(1)e^(r_(+)x)+c_(2)e^(r_(-)x ))

   Two complex roots. It follows from the fundamental theorem of algebra that solutions to polynomial equations with real coefficients have roots that are real or form conjugate pairs. Therefore, if the complex number r = α + i β (\displaystyle r=\alpha +i\beta ) is the root of the characteristic equation, then r ∗ = α − i β (\displaystyle r^(*)=\alpha -i\beta ) is also the root of this equation. Thus, the solution can be written in the form c 1 e (α + i β) x + c 2 e (α − i β) x , (\displaystyle c_(1)e^((\alpha +i\beta)x)+c_(2)e^( (\alpha -i\beta)x),) however, this is a complex number and is undesirable in solving practical problems.

   • Instead, you can use Euler formula e i x = cos ⁡ x + i sin ⁡ x (\displaystyle e^(ix)=\cos x+i\sin x), which allows us to write the solution in the form trigonometric functions:
     • e α x (c 1 cos ⁡ β x + i c 1 sin ⁡ β x + c 2 cos ⁡ β x − i c 2 sin ⁡ β x) (\displaystyle e^(\alpha x)(c_(1)\cos \ beta x+ic_(1)\sin \beta x+c_(2)\cos \beta x-ic_(2)\sin \beta x))
   • Now you can instead of constant c 1 + c 2 (\displaystyle c_(1)+c_(2)) write down c 1 (\displaystyle c_(1)), and the expression i (c 1 − c 2) (\displaystyle i(c_(1)-c_(2))) replaced by c 2 . (\displaystyle c_(2).) After that we get the following solution:
     • y (x) = e α x (c 1 cos ⁡ β x + c 2 sin ⁡ β x) (\displaystyle y(x)=e^(\alpha x)(c_(1)\cos \beta x+c_ (2)\sin \beta x))
   • There is another way to write the solution in terms of amplitude and phase, which is better suited for physical problems.
   • Example 2.1. Let us find the solution of the differential equation given below with given initial conditions. For this, it is necessary to take the obtained solution, as well as its derivative, and substitute them into the initial conditions, which will allow us to determine arbitrary constants.
     • d 2 x d t 2 + 3 d x d t + 10 x = 0 , x (0) = 1 , x ′ (0) = − 1 (\displaystyle (\frac ((\mathrm (d) )^(2)x)(( \mathrm (d) )t^(2)))+3(\frac ((\mathrm (d) )x)((\mathrm (d) )t))+10x=0,\quad x(0) =1,\ x"(0)=-1)
     • r 2 + 3 r + 10 = 0 , r ± = − 3 ± 9 − 40 2 = − 3 2 ± 31 2 i (\displaystyle r^(2)+3r+10=0,\quad r_(\pm ) =(\frac (-3\pm (\sqrt (9-40)))(2))=-(\frac (3)(2))\pm (\frac (\sqrt (31))(2) )i)
     • x (t) = e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(c_(1 )\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right))
     • x (0) = 1 = c 1 (\displaystyle x(0)=1=c_(1))
     • x ′ (t) = − 3 2 e − 3 t / 2 (c 1 cos ⁡ 31 2 t + c 2 sin ⁡ 31 2 t) + e − 3 t / 2 (− 31 2 c 1 sin ⁡ 31 2 t + 31 2 c 2 cos ⁡ 31 2 t) (\displaystyle (\begin(aligned)x"(t)&=-(\frac (3)(2))e^(-3t/2)\left(c_ (1)\cos (\frac (\sqrt (31))(2))t+c_(2)\sin (\frac (\sqrt (31))(2))t\right)\\&+e ^(-3t/2)\left(-(\frac (\sqrt (31))(2))c_(1)\sin (\frac (\sqrt (31))(2))t+(\frac ( \sqrt (31))(2))c_(2)\cos (\frac (\sqrt (31))(2))t\right)\end(aligned)))
     • x ′ (0) = − 1 = − 3 2 c 1 + 31 2 c 2 , c 2 = 1 31 (\displaystyle x"(0)=-1=-(\frac (3)(2))c_( 1)+(\frac (\sqrt (31))(2))c_(2),\quad c_(2)=(\frac (1)(\sqrt (31))))
     • x (t) = e − 3 t / 2 (cos ⁡ 31 2 t + 1 31 sin ⁡ 31 2 t) (\displaystyle x(t)=e^(-3t/2)\left(\cos (\frac (\sqrt (31))(2))t+(\frac (1)(\sqrt (31)))\sin (\frac (\sqrt (31))(2))t\right))
   
   Solving differential equations of the nth order with constant coefficients (recorded by Intuit - National Open University).

2. Downgrading order. Order reduction is a method for solving differential equations when one linearly independent solution is known. This method consists in lowering the order of the equation by one, which allows the equation to be solved using the methods described in the previous section. Let the solution be known. The main idea of ​​lowering the order is to find a solution in the form below, where it is necessary to define the function v (x) (\displaystyle v(x)), substituting it into the differential equation and finding v(x). (\displaystyle v(x).) Let's consider how order reduction can be used to solve a differential equation with constant coefficients and multiple roots.

   
   

   Multiple roots homogeneous differential equation with constant coefficients. Recall that a second-order equation must have two linearly independent solutions. If the characteristic equation has multiple roots, the set of solutions not forms a space since these solutions are linearly dependent. In this case, order reduction must be used to find a second linearly independent solution.

   • Let the characteristic equation have multiple roots r (\displaystyle r). We assume that the second solution can be written as y (x) = e r x v (x) (\displaystyle y(x)=e^(rx)v(x)), and substitute it into the differential equation. In this case, most of the terms, with the exception of the term with the second derivative of the function v , (\displaystyle v,) will be reduced.
     • v ″ (x) e r x = 0 (\displaystyle v""(x)e^(rx)=0)
   • Example 2.2. Given the following equation, which has multiple roots r = − 4. (\displaystyle r=-4.) When substituting, most of the terms are cancelled.
     • d 2 y d x 2 + 8 d y d x + 16 y = 0 (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+8( \frac ((\mathrm (d) )y)((\mathrm (d) )x))+16y=0)
     • y = v (x) e − 4 x y ′ = v ′ (x) e − 4 x − 4 v (x) e − 4 x y ″ = v ″ (x) e − 4 x − 8 v ′ (x) e − 4 x + 16 v (x) e − 4 x (\displaystyle (\begin(aligned)y&=v(x)e^(-4x)\\y"&=v"(x)e^(-4x )-4v(x)e^(-4x)\\y""&=v""(x)e^(-4x)-8v"(x)e^(-4x)+16v(x)e^ (-4x)\end(aligned)))
     • v ″ e − 4 x − 8 v ′ e − 4 x + 16 v e − 4 x + 8 v ′ e − 4 x − 32 v e − 4 x + 16 v e − 4 x = 0 (\displaystyle (\begin(aligned )v""e^(-4x)&-(\cancel (8v"e^(-4x)))+(\cancel (16ve^(-4x)))\\&+(\cancel (8v"e ^(-4x)))-(\cancel (32ve^(-4x)))+(\cancel (16ve^(-4x)))=0\end(aligned)))
   • Like our ansatz for a differential equation with constant coefficients, in this case only the second derivative can be equal to zero. We integrate twice and obtain the desired expression for v (\displaystyle v):
     • v (x) = c 1 + c 2 x (\displaystyle v(x)=c_(1)+c_(2)x)
   • Then the general solution of a differential equation with constant coefficients, if the characteristic equation has multiple roots, can be written in the following form. For convenience, you can remember that to obtain linear independence, it is enough to simply multiply the second term by x (\displaystyle x). This set of solutions is linearly independent, and thus we have found all solutions to this equation.
     • y (x) = (c 1 + c 2 x) e r x (\displaystyle y(x)=(c_(1)+c_(2)x)e^(rx))

   D 2 y d x 2 + p (x) d y d x + q (x) y = 0. (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^( 2)))+p(x)(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+q(x)y=0.) Order reduction is applicable if the solution is known y 1 (x) (\displaystyle y_(1)(x)), which can be found or given in the problem statement.

   • We are looking for a solution in the form y (x) = v (x) y 1 (x) (\displaystyle y(x)=v(x)y_(1)(x)) and plug it into this equation:
     • v ″ y 1 + 2 v ′ y 1 ′ + p (x) v ′ y 1 + v (y 1 ″ + p (x) y 1 ′ + q (x)) = 0 (\displaystyle v""y_( 1)+2v"y_(1)"+p(x)v"y_(1)+v(y_(1)""+p(x)y_(1)"+q(x))=0)
   • Because the y 1 (\displaystyle y_(1)) is a solution to the differential equation, all terms with v (\displaystyle v) are shrinking. As a result, it remains first order linear equation. To see this more clearly, let us change the variables w (x) = v′ (x) (\displaystyle w(x)=v"(x)):
     • y 1 w ′ + (2 y 1 ′ + p (x) y 1) w = 0 (\displaystyle y_(1)w"+(2y_(1)"+p(x)y_(1))w=0 )
     • w (x) = exp ⁡ (∫ (2 y 1 ′ (x) y 1 (x) + p (x)) d x) (\displaystyle w(x)=\exp \left(\int \left((\ frac (2y_(1)"(x))(y_(1)(x)))+p(x)\right)(\mathrm (d) )x\right))
     • v (x) = ∫ w (x) d x (\displaystyle v(x)=\int w(x)(\mathrm (d) )x)
   • If the integrals can be calculated, we get the general solution as a combination of elementary functions. Otherwise, the solution can be left in integral form.

3. Cauchy-Euler equation. The Cauchy-Euler equation is an example of a second-order differential equation with variables coefficients, which has exact solutions. This equation is used in practice, for example, to solve the Laplace equation in spherical coordinates.

   X 2 d 2 y d x 2 + a x d y d x + b y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2) ))+ax(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=0)
   

   Characteristic equation. As you can see, in this differential equation, each term contains a power factor, the degree of which is equal to the order of the corresponding derivative.

   • Thus, one can try to look for a solution in the form y (x) = x n , (\displaystyle y(x)=x^(n),) where to define n (\displaystyle n), just as we were looking for a solution in the form of an exponential function for a linear differential equation with constant coefficients. After differentiation and substitution, we get
     • x n (n 2 + (a − 1) n + b) = 0 (\displaystyle x^(n)(n^(2)+(a-1)n+b)=0)
   • To use the characteristic equation, we must assume that x ≠ 0 (\displaystyle x\neq 0). Dot x = 0 (\displaystyle x=0) called regular singular point differential equation. Such points are important when solving differential equations using power series. This equation has two roots, which can be different and real, multiple or complex conjugate.
     • n ± = 1 − a ± (a − 1) 2 − 4 b 2 (\displaystyle n_(\pm )=(\frac (1-a\pm (\sqrt ((a-1)^(2)-4b )))(2)))

   Two different real roots. If the roots n ± (\displaystyle n_(\pm )) are real and different, then the solution of the differential equation has the following form:

   • y (x) = c 1 x n + + c 2 x n − (\displaystyle y(x)=c_(1)x^(n_(+))+c_(2)x^(n_(-)))

   Two complex roots. If the characteristic equation has roots n ± = α ± β i (\displaystyle n_(\pm )=\alpha \pm \beta i), the solution is a complex function.

   • To transform the solution into a real function, we make a change of variables x = e t , (\displaystyle x=e^(t),) that is t = ln ⁡ x , (\displaystyle t=\ln x,) and use the Euler formula. Similar actions were performed earlier when defining arbitrary constants.
     • y (t) = e α t (c 1 e β i t + c 2 e − β i t) (\displaystyle y(t)=e^(\alpha t)(c_(1)e^(\beta it)+ c_(2)e^(-\beta it)))
   • Then the general solution can be written as
     • y (x) = x α (c 1 cos ⁡ (β ln ⁡ x) + c 2 sin ⁡ (β ln ⁡ x)) (\displaystyle y(x)=x^(\alpha )(c_(1)\ cos(\beta \ln x)+c_(2)\sin(\beta \ln x)))

   Multiple roots. To obtain a second linearly independent solution, it is necessary to reduce the order again.

   • It takes quite a bit of computation, but the principle is the same: we substitute y = v (x) y 1 (\displaystyle y=v(x)y_(1)) into an equation whose first solution is y 1 (\displaystyle y_(1)). After reductions, the following equation is obtained:
     • v ″ + 1 x v ′ = 0 (\displaystyle v""+(\frac (1)(x))v"=0)
   • This is a first order linear equation with respect to v′ (x) . (\displaystyle v"(x).) His solution is v (x) = c 1 + c 2 ln ⁡ x . (\displaystyle v(x)=c_(1)+c_(2)\ln x.) Thus, the solution can be written in the following form. It's pretty easy to remember - to get the second linearly independent solution, you just need an additional term with ln ⁡ x (\displaystyle \ln x).
     • y (x) = x n (c 1 + c 2 ln ⁡ x) (\displaystyle y(x)=x^(n)(c_(1)+c_(2)\ln x))

4. Inhomogeneous linear differential equations with constant coefficients. Nonhomogeneous equations have the form L [ y (x) ] = f (x) , (\displaystyle L=f(x),) where f (x) (\displaystyle f(x))- so-called free member. According to the theory of differential equations, the general solution of this equation is a superposition private decision y p (x) (\displaystyle y_(p)(x)) and additional solution y c (x) . (\displaystyle y_(c)(x).) However, in this case, a particular solution does not mean a solution given by the initial conditions, but rather a solution that is due to the presence of inhomogeneity (free member). The complementary solution is the solution of the corresponding homogeneous equation in which f (x) = 0. (\displaystyle f(x)=0.) The general solution is a superposition of these two solutions, because L [ y p + y c ] = L [ y p ] + L [ y c ] = f (x) (\displaystyle L=L+L=f(x)), and since L [ y c ] = 0 , (\displaystyle L=0,) such a superposition is indeed a general solution.

   D 2 y d x 2 + a d y d x + b y = f (x) (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )x^(2)))+a (\frac ((\mathrm (d) )y)((\mathrm (d) )x))+by=f(x))
   

   Method uncertain coefficients. The method of indeterminate coefficients is used in cases where the free term is a combination of exponential, trigonometric, hyperbolic or power functions. Only these functions are guaranteed to have a finite number of linearly independent derivatives. In this section, we will find a particular solution to the equation.

   • Compare the terms in f (x) (\displaystyle f(x)) with terms in ignoring constant factors. Three cases are possible.
     • There are no identical members. In this case, a particular solution y p (\displaystyle y_(p)) will be a linear combination of terms from y p (\displaystyle y_(p))
     • f (x) (\displaystyle f(x)) contains member x n (\displaystyle x^(n)) and a member from y c , (\displaystyle y_(c),) where n (\displaystyle n) is zero or a positive integer, and this term corresponds to a single root of the characteristic equation. In this case y p (\displaystyle y_(p)) will consist of a combination of the function x n + 1 h (x) , (\displaystyle x^(n+1)h(x),) its linearly independent derivatives, as well as other terms f (x) (\displaystyle f(x)) and their linearly independent derivatives.
     • f (x) (\displaystyle f(x)) contains member h (x) , (\displaystyle h(x),) which is a work x n (\displaystyle x^(n)) and a member from y c , (\displaystyle y_(c),) where n (\displaystyle n) is equal to 0 or a positive integer, and this term corresponds to multiple root of the characteristic equation. In this case y p (\displaystyle y_(p)) is a linear combination of the function x n + s h (x) (\displaystyle x^(n+s)h(x))(where s (\displaystyle s)- multiplicity of the root) and its linearly independent derivatives, as well as other members of the function f (x) (\displaystyle f(x)) and its linearly independent derivatives.
   • Let's write down y p (\displaystyle y_(p)) as a linear combination of the above terms. Thanks to these coefficients in a linear combination this method called the method of indeterminate coefficients. Upon the appearance of those contained in y c (\displaystyle y_(c)) their members can be discarded due to the presence of arbitrary constants in y c . (\displaystyle y_(c).) After that we substitute y p (\displaystyle y_(p)) into an equation and equate like terms.
   • We determine the coefficients. At this stage, the system algebraic equations, which can usually be solved without any problems. The solution of this system makes it possible to obtain y p (\displaystyle y_(p)) and thereby solve the equation.
   • Example 2.3. Consider an inhomogeneous differential equation whose free term contains a finite number of linearly independent derivatives. A particular solution of such an equation can be found by the method of indefinite coefficients.
     • d 2 y d t 2 + 6 y = 2 e 3 t − cos ⁡ 5 t (\displaystyle (\frac ((\mathrm (d) )^(2)y)((\mathrm (d) )t^(2) ))+6y=2e^(3t)-\cos 5t)
     • y c (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t (\displaystyle y_(c)(t)=c_(1)\cos (\sqrt (6))t+c_(2)\sin (\sqrt(6))t)
     • y p (t) = A e 3 t + B cos ⁡ 5 t + C sin ⁡ 5 t (\displaystyle y_(p)(t)=Ae^(3t)+B\cos 5t+C\sin 5t)
     • 9 A e 3 t − 25 B cos ⁡ 5 t − 25 C sin ⁡ 5 t + 6 A e 3 t + 6 B cos ⁡ 5 t + 6 C sin ⁡ 5 t = 2 e 3 t − cos ⁡ 5 t ( \displaystyle (\begin(aligned)9Ae^(3t)-25B\cos 5t&-25C\sin 5t+6Ae^(3t)\\&+6B\cos 5t+6C\sin 5t=2e^(3t)-\ cos 5t\end(aligned)))
     • ( 9 A + 6 A = 2 , A = 2 15 − 25 B + 6 B = − 1 , B = 1 19 − 25 C + 6 C = 0 , C = 0 (\displaystyle (\begin(cases)9A+ 6A=2,&A=(\dfrac (2)(15))\\-25B+6B=-1,&B=(\dfrac (1)(19))\\-25C+6C=0,&C=0 \end(cases)))
     • y (t) = c 1 cos ⁡ 6 t + c 2 sin ⁡ 6 t + 2 15 e 3 t + 1 19 cos ⁡ 5 t (\displaystyle y(t)=c_(1)\cos (\sqrt (6 ))t+c_(2)\sin (\sqrt (6))t+(\frac (2)(15))e^(3t)+(\frac (1)(19))\cos 5t)

   Lagrange method. The Lagrange method, or the method of variation of arbitrary constants, is a more general method for solving inhomogeneous differential equations, especially in cases where the free term does not contain a finite number of linearly independent derivatives. For example, with free members tan ⁡ x (\displaystyle \tan x) or x − n (\displaystyle x^(-n)) to find a particular solution, it is necessary to use the Lagrange method. The Lagrange method can even be used to solve differential equations with variable coefficients, although in this case, with the exception of the Cauchy-Euler equation, it is less often used, since the additional solution is usually not expressed in terms of elementary functions.

   • Let's assume that the solution has the following form. Its derivative is given in the second line.
     • y (x) = v 1 (x) y 1 (x) + v 2 (x) y 2 (x) (\displaystyle y(x)=v_(1)(x)y_(1)(x)+v_ (2)(x)y_(2)(x))
     • y ′ = v 1 ′ y 1 + v 1 y 1 ′ + v 2 ′ y 2 + v 2 y 2 ′ (\displaystyle y"=v_(1)"y_(1)+v_(1)y_(1) "+v_(2)"y_(2)+v_(2)y_(2)")
   • Since the proposed solution contains two unknown quantities, it is necessary to impose additional condition. We choose this additional condition in the following form:
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 (\displaystyle v_(1)"y_(1)+v_(2)"y_(2)=0)
     • y ′ = v 1 y 1 ′ + v 2 y 2 ′ (\displaystyle y"=v_(1)y_(1)"+v_(2)y_(2)")
     • y ″ = v 1 ′ y 1 ′ + v 1 y 1 ″ + v 2 ′ y 2 ′ + v 2 y 2 ″ (\displaystyle y""=v_(1)"y_(1)"+v_(1) y_(1)""+v_(2)"y_(2)"+v_(2)y_(2)"")
   • Now we can get the second equation. After substituting and redistributing members, you can group together members with v 1 (\displaystyle v_(1)) and members from v 2 (\displaystyle v_(2)). These terms are canceled because y 1 (\displaystyle y_(1)) and y 2 (\displaystyle y_(2)) are solutions of the corresponding homogeneous equation. As a result, we obtain the following system of equations
     • v 1 ′ y 1 + v 2 ′ y 2 = 0 v 1 ′ y 1 ′ + v 2 ′ y 2 ′ = f (x) (\displaystyle (\begin(aligned)v_(1)"y_(1)+ v_(2)"y_(2)&=0\\v_(1)"y_(1)"+v_(2)"y_(2)"&=f(x)\\\end(aligned)))
   • This system can be transformed into a matrix equation of the form A x = b , (\displaystyle A(\mathbf (x) )=(\mathbf (b) ),) whose solution is x = A − 1 b . (\displaystyle (\mathbf (x) )=A^(-1)(\mathbf (b) ).) For matrix 2 × 2 (\displaystyle 2\times 2) inverse matrix is found by dividing by the determinant, permuting the diagonal elements, and changing the sign of the off-diagonal elements. In fact, the determinant of this matrix is ​​a Wronskian.
     • (v 1 ′ v 2 ′) = 1 W (y 2 ′ − y 2 − y 1 ′ y 1) (0 f (x)) (\displaystyle (\begin(pmatrix)v_(1)"\\v_( 2)"\end(pmatrix))=(\frac (1)(W))(\begin(pmatrix)y_(2)"&-y_(2)\\-y_(1)"&y_(1)\ end(pmatrix))(\begin(pmatrix)0\\f(x)\end(pmatrix)))
   • Expressions for v 1 (\displaystyle v_(1)) and v 2 (\displaystyle v_(2)) are listed below. As in the order reduction method, in this case an arbitrary constant appears during integration, which includes an additional solution in the general solution of the differential equation.
     • v 1 (x) = − ∫ 1 W f (x) y 2 (x) d x (\displaystyle v_(1)(x)=-\int (\frac (1)(W))f(x)y_( 2)(x)(\mathrm (d) )x)
     • v 2 (x) = ∫ 1 W f (x) y 1 (x) d x (\displaystyle v_(2)(x)=\int (\frac (1)(W))f(x)y_(1) (x)(\mathrm (d) )x)
   
   Lecture of the National Open University Intuit entitled "Linear differential equations of the n-th order with constant coefficients".

Practical use

Differential equations establish a relationship between a function and one or more of its derivatives. Since such relationships are so common, differential equations have found wide application in a wide variety of areas, and since we live in four dimensions, these equations are often differential equations in private derivatives. This section discusses some of the most important equations of this type.

• Exponential growth and decay. radioactive decay. Compound interest. Speed chemical reactions. The concentration of drugs in the blood. Unlimited population growth. Newton-Richmann law. In the real world, there are many systems in which the rate of growth or decay at any given time is proportional to the amount at that time, or can be well approximated by a model. This is because the solution to this differential equation, the exponential function, is one of the most important functions in mathematics and other sciences. More generally, under controlled population growth, the system may include additional terms that limit growth. In the equation below, the constant k (\displaystyle k) can be either greater or less than zero.
  • d y d x = k x (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=kx)
• Harmonic vibrations. In both classical and quantum mechanics, the harmonic oscillator is one of the most important physical systems due to its simplicity and wide application for approximating more complex systems such as a simple pendulum. In classical mechanics harmonic vibrations are described by an equation that relates the position material point with its acceleration by Hooke's law. In this case, damping and driving forces can also be taken into account. In the expression below x ˙ (\displaystyle (\dot (x)))- time derivative of x , (\displaystyle x,) β (\displaystyle \beta ) is a parameter that describes the damping force, ω 0 (\displaystyle \omega _(0))- angular frequency of the system, F (t) (\displaystyle F(t))- time dependent driving force. The harmonic oscillator is also present in electromagnetic oscillatory circuits, where it can be implemented with greater accuracy than in mechanical systems.
  • x ¨ + 2 β x ˙ + ω 0 2 x = F (t) (\displaystyle (\ddot (x))+2\beta (\dot (x))+\omega _(0)^(2)x =F(t))
• Bessel equation. The Bessel differential equation is used in many areas of physics, including solving wave equation, Laplace equations and Schrödinger equations, especially in the presence of cylindrical or spherical symmetry. This second-order differential equation with variable coefficients is not a Cauchy-Euler equation, so its solutions cannot be written as elementary functions. The solutions of the Bessel equation are the Bessel functions, which are well studied due to the fact that they are used in many areas. In the expression below α (\displaystyle \alpha ) is a constant that matches order Bessel functions.
  • x 2 d 2 y d x 2 + x d y d x + (x 2 − α 2) y = 0 (\displaystyle x^(2)(\frac ((\mathrm (d) )^(2)y)((\mathrm (d ) )x^(2)))+x(\frac ((\mathrm (d) )y)((\mathrm (d) )x))+(x^(2)-\alpha ^(2)) y=0)
• Maxwell's equations. Along with the Lorentz force, Maxwell's equations form the basis of classical electrodynamics. These are four partial differential equations for the electric E (r , t) (\displaystyle (\mathbf (E) )((\mathbf (r) ),t)) and magnetic B (r , t) (\displaystyle (\mathbf (B) )((\mathbf (r) ),t)) fields. In the expressions below ρ = ρ (r , t) (\displaystyle \rho =\rho ((\mathbf (r) ),t))- charge density, J = J (r , t) (\displaystyle (\mathbf (J) )=(\mathbf (J) )((\mathbf (r) ),t)) is the current density, and ϵ 0 (\displaystyle \epsilon _(0)) and μ 0 (\displaystyle \mu _(0)) are the electric and magnetic constants, respectively.
  • ∇ ⋅ E = ρ ϵ 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × B = μ 0 J + μ 0 ϵ 0 ∂ E ∂ t (\displaystyle (\begin(aligned)\nabla \cdot (\mathbf (E) )&=(\frac (\rho )(\epsilon _(0)))\\\nabla \cdot (\mathbf (B) )&=0\\\nabla \times (\mathbf (E) )&=-(\frac (\partial (\mathbf (B) ))(\partial t))\\\nabla \times (\mathbf (B) )&=\mu _(0)(\ mathbf (J) )+\mu _(0)\epsilon _(0)(\frac (\partial (\mathbf (E) ))(\partial t))\end(aligned)))
• Schrödinger equation. In quantum mechanics, the Schrödinger equation is the basic equation of motion that describes the movement of particles in accordance with the change in the wave function Ψ = Ψ (r , t) (\displaystyle \Psi =\Psi ((\mathbf (r) ),t)) with time. The equation of motion is described by the behavior Hamiltonian H ^ (\displaystyle (\hat(H))) - operator, which describes the energy of the system. One of the widely famous examples the Schrödinger equation in physics is the equation for one non-relativistic particle, which is subject to the potential V (r , t) (\displaystyle V((\mathbf (r) ),t)). Many systems are described by the time-dependent Schrödinger equation, with the equation on the left side E Ψ , (\displaystyle E\Psi ,) where E (\displaystyle E) is the energy of the particle. In the expressions below ℏ (\displaystyle \hbar ) is the reduced Planck constant.
  • i ℏ ∂ Ψ ∂ t = H ^ Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=(\hat (H))\Psi )
  • i ℏ ∂ Ψ ∂ t = (− ℏ 2 2 m ∇ 2 + V (r , t)) Ψ (\displaystyle i\hbar (\frac (\partial \Psi )(\partial t))=\left(- (\frac (\hbar ^(2))(2m))\nabla ^(2)+V((\mathbf (r) ),t)\right)\Psi )
• wave equation. It is impossible to imagine physics and technology without waves, they are present in all types of systems. In general, waves are described by the equation below, in which u = u (r , t) (\displaystyle u=u((\mathbf (r) ),t)) is the desired function, and c (\displaystyle c)- experimentally determined constant. d'Alembert was the first to discover that for the one-dimensional case the solution to the wave equation is any function with argument x − c t (\displaystyle x-ct), which describes an arbitrary wave propagating to the right. The general solution for the one-dimensional case is a linear combination of this function with a second function with an argument x + c t (\displaystyle x+ct), which describes a wave propagating to the left. This solution is presented in the second line.
  • ∂ 2 u ∂ t 2 = c 2 ∇ 2 u (\displaystyle (\frac (\partial ^(2)u)(\partial t^(2)))=c^(2)\nabla ^(2)u )
  • u (x , t) = f (x − c t) + g (x + c t) (\displaystyle u(x,t)=f(x-ct)+g(x+ct))
• Navier-Stokes equations. The Navier-Stokes equations describe the movement of fluids. Since fluids are present in virtually every field of science and technology, these equations are extremely important for weather prediction, aircraft design, ocean currents, and many other applications. The Navier-Stokes equations are non-linear partial differential equations, and in most cases it is very difficult to solve them, since the non-linearity leads to turbulence, and in order to obtain a stable solution by numerical methods, partitioning into very small cells is necessary, which requires significant computing power. For practical purposes in hydrodynamics, methods such as time averaging are used to model turbulent flows. Challenges are even more basic questions, such as the existence and uniqueness of solutions for non-linear partial differential equations, and proving the existence and uniqueness of a solution for the Navier-Stokes equations in three dimensions is among math problems millennium. Below are the incompressible fluid flow equation and the continuity equation.
  • ∂ u ∂ t + (u ⋅ ∇) u − ν ∇ 2 u = − ∇ h , ∂ ρ ∂ t + ∇ ⋅ (ρ u) = 0 (\displaystyle (\frac (\partial (\mathbf (u) ) )(\partial t))+((\mathbf (u) )\cdot \nabla)(\mathbf (u) )-\nu \nabla ^(2)(\mathbf (u) )=-\nabla h, \quad (\frac (\partial \rho )(\partial t))+\nabla \cdot (\rho (\mathbf (u) ))=0)

• Many differential equations simply cannot be solved by the above methods, especially those mentioned in the last section. This applies to cases where the equation contains variable odds and is not a Cauchy-Euler equation, or when the equation is non-linear, except in a few very rare cases. However, the above methods allow you to solve many important differential equations that are often encountered in various fields of science.
• Unlike differentiation, which allows you to find the derivative of any function, the integral of many expressions cannot be expressed in elementary functions. Therefore, do not waste time trying to calculate the integral where it is impossible. Look at the table of integrals. If the solution of a differential equation cannot be expressed in terms of elementary functions, sometimes it can be represented in integral form, and in this case it does not matter whether this integral can be calculated analytically.

Warnings

• Appearance differential equation can be misleading. For example, below are two first-order differential equations. The first equation is easily solved using the methods described in this article. At first glance, a minor change y (\displaystyle y) on the y 2 (\displaystyle y^(2)) in the second equation makes it non-linear and becomes very difficult to solve.
  • d y d x = x 2 + y (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y)
  • d y d x = x 2 + y 2 (\displaystyle (\frac ((\mathrm (d) )y)((\mathrm (d) )x))=x^(2)+y^(2))