Types of general solution of differential equations. Basic concepts and definitions of differential equations

differential equation an equation is called that relates the independent variable x, the desired function y=f(x) and its derivatives y",y"",\ldots,y^((n)), i.e., an equation of the form

F(x,y,y",y"",\ldots,y^((n)))=0.

If the desired function y \u003d y (x) is a function of one independent variable x, the differential equation is called ordinary; for example,

\mathsf(1))~\frac(dy)(dx)+xy=0, \quad \mathsf(2))~y""+y"+x=\cos(x), \quad \mathsf(3 ))~(x^2-y^2)\,dx-(x+y)\,dy=0.

When the desired function y is a function of two or more independent variables, for example, if y=y(x,t) , then an equation of the form

F\!\left(x,t,y,\frac(\partial(y))(\partial(x)),\frac(\partial(y))(\partial(t)),\ldots,\ frac(\partial^m(y))(\partial(x^k)\partial(t^l))\right)=0

is called a partial differential equation. Here k,l are non-negative integers such that k+l=m ; for example

\frac(\partial(y))(\partial(t))-\frac(\partial(y))(\partial(x))=0, \quad \frac(\partial(y))(\partial (t))=\frac(\partial^2y)(\partial(x^2)).

The order of the differential equation is the order of the highest derivative in the equation. For example, the differential equation y"+xy=e^x is a first-order equation, the differential equation y""+p(x)y=0, where p(x) is a known function, is a second-order equation; the differential equation y^( (9))-xy""=x^2 - 9th order equation.

By solving the differential equation nth order on the interval (a,b) is a function y=\varphi(x) defined on the interval (a,b) together with its derivatives up to the nth order inclusive, and such that the substitution of the function y=\varphi (x) into a differential equation turns the latter into an identity in x on (a,b) . For example, the function y=\sin(x)+\cos(x) is a solution of the equation y""+y=0 on the interval (-\infty,+\infty). Indeed, differentiating the function twice, we have

y"=\cos(x)-\sin(x), \quad y""=-\sin(x)-\cos(x).

Substituting the expressions y"" and y into the differential equation, we obtain the identity

-\sin(x)-\cos(x)+\sin(x)+\cos(x)\equiv0

The graph for solving a differential equation is called integral curve this equation.

General view of the first order equation

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

If equation (1) can be solved with respect to y" , then we get first order equation solved with respect to the derivative.

y"=f(x,y).

The Cauchy problem is the problem of finding a solution y=y(x) of the equation y"=f(x,y) that satisfies the initial condition y(x_0)=y_0 (another notation y|_(x=x_0)=y_0 ).

Geometrically, this means that we are looking for an integral curve passing through a given
point M_0(x_0,y_0) of the xOy plane (Fig. 1).

Existence and uniqueness theorem for the solution of the Cauchy problem

Let a differential equation be given y"=f(x,y) , where the function f(x,y) is defined in some region D of the xOy plane containing the point (x_0,y_0) . If the function f(x,y) satisfies the conditions

a) f(x,y) is a continuous function of two variables x and y in the domain D ;

b) f(x,y) has a partial derivative bounded in the region D , then there is an interval (x_0-h,x_0+h) on which there is a unique solution y=\varphi(x) of this equation that satisfies the condition y(x_0 )=y_0 .

The theorem gives sufficient conditions for the existence of a unique solution to the Cauchy problem for the equation y"=f(x,y) , but these conditions are not necessary. Namely, there can be a unique solution of the equation y"=f(x,y) that satisfies the condition y(x_0)=y_0 , although conditions a) or b) or both are not satisfied at the point (x_0,y_0).

Consider examples.

1. y"=\frac(1)(y^2) . Here f(x,y)=\frac(1)(y^2),~\frac(\partial(f))(\partial(y))=-\frac(2)(y^3). At points (x_0,0) of the Ox axis, conditions a) and b) are not satisfied (function f(x,y) and its partial derivative \frac(\partial(f))(\partial(y)) are discontinuous on the Ox axis and unbounded for y\to0 ), but a single integral curve passes through each point of the Ox axis y=\sqrt(3(x-x_0))(Fig. 2).

2. y"=xy+e^(-y) . The right side of the equation f(x,y)=xy+e^(-y) and its partial derivative \frac(\partial(f))(\partial(y))=x-e^(-y) are continuous in x and y at all points of the plane xOy . By virtue of the existence and uniqueness theorem, the region in which the given equation has a unique solution
is the entire plane xOy .

3. y"=\frac(3)(2)\sqrt(y^2). Right hand side of the equation f(x,y)=\frac(3)(2)\sqrt(y^2) is defined and continuous at all points of the xOy plane. Partial derivative \frac(\partial(f))(\partial(y))=\frac(1)(\sqrt(y)) goes to infinity at y=0 , i.e. on the axis Ox , so that condition b) of the existence and uniqueness theorem is violated for y=0. Consequently, at the points of the Ox axis, uniqueness can be violated. It is easy to check that the function is a solution of the given equation. In addition, the equation has an obvious solution y\equiv0 . Thus, at least two integral lines pass through each point of the Ox axis and, consequently, uniqueness is indeed violated at the points of this axis (Fig. 3).

The integral lines of this equation will also be lines composed of pieces of cubic parabolas y=\frac((x+c)^3)(8) and segments of the Ox axis, for example, ABOC_1, ABB_2C_2, A_2B_2x, etc., so that an infinite set of integral lines passes through each point of the Ox axis.

Lipschitz condition

Comment. Derivative Bounded Condition \partial(f)/\partial(y), appearing in the existence and uniqueness theorem for the solution of the Cauchy problem, can be somewhat weakened and replaced by the so-called Lipschitz condition.

It is said that a function f(x,y) defined in some domain D satisfies the Lipschitz condition on y in D if there exists such a constant L ( Lipschitz constant) that for any y_1,y_2 from D and any x from D the inequality

|f(x,y_2)-f(x,y_1)| \leqslant L|y_2-y_1|.

Existence in the domain D of a bounded derivative \frac(\partial(f))(\partial(y)) is sufficient for the function f(x,y) to satisfy the Lipschitz condition in D. On the contrary, the Lipschitz condition does not imply the boundedness condition \frac(\partial(f))(\partial(y)); the latter may not even exist. For example, for the equation y"=2|y|\cos(x) the function f(x,y)=2|y|\cos(x) not differentiable with respect to y at a point (x_0,0),x_0\ne\frac(\pi)(2)+k\pi,k\in\mathbb(Z), but the Lipschitz condition is satisfied in a neighborhood of this point. Indeed,

(|f(x,y_2)-f(x,y_1)|=L|2|y_2|\cos(x)-2|y_1|\cos(x)|=2|\cos(x)|\, ||y_2|-|y_1||\leqslant2|y_2-y_1|.)

because the |\cos(x)|\leqslant1, a ||y_2|-|y_1||\leqslant|y_2-y_1|. Thus, the Lipschitz condition is satisfied with the constant L=2 .

Theorem. If the function f(x,y) is continuous and satisfies the Lipschitz condition on y in the domain D , then the Cauchy problem

\frac(dy)(dx)=f(x,y), \quad y|_(x=x_0)=y_0, \quad (x_0,y_0)\in(D).

has a unique solution.

The Lipschitz condition is essential for the uniqueness of the solution of the Cauchy problem. As an example, consider the equation

\frac(dy)(dx)=\begin(cases)\dfrac(4x^3y)(x^4+y^4),&x^2+y^2>0,\\0,&x=y=0 .\end(cases)

It is easy to see that the function f(x, y) is continuous; on the other hand,

f(x,Y)-f(x,y)=\frac(4x^3(x^4+yY))((x^4+y^2)(x^4+Y^2))(Y-y ).

If a y=\alpha x^2,~Y=\beta x^2, then

|f(x,Y)-f(x,y)|=\frac(4)(|x|)\frac(1-\alpha\beta)((1+\alpha^2)(1+\beta ^2))|Y-y|,

and the Lipschitz condition is not satisfied in any region containing the origin O(0,0) , since the factor at |Y-y| turns out to be unbounded for x\to0 .

This differential equation admits a solution y=C^2-\sqrt(x^4+C^4), where C is an arbitrary constant. This shows that there is an infinite set of solutions that satisfy the initial condition y(0)=0.

General solution differential equation (2) is called the function

y=\varphi(x,C),

depending on one arbitrary constant C , and such that

1) it satisfies equation (2) for any admissible values of the constant C;

2) whatever the initial condition

\Bigl.(y)\Bigr|_(x=x_0)=y_0,

one can choose such a value C_0 of the constant C that the solution y=\varphi(x,C_0) will satisfy the given initial condition (4). It is assumed that the point (x_0,y_0) belongs to the region where the conditions for the existence and uniqueness of the solution are satisfied.

Private decision differential equation (2) is called the solution obtained from common solution(3) for some specific value of an arbitrary constant C .

Example 1. Check that the function y=x+C is a general solution of the differential equation y"=1 and find a particular solution that satisfies the initial condition y|_(x=0)=0. Give a geometric interpretation of the result.

Solution. The function y=x+C satisfies this equation for any values of an arbitrary constant C . Indeed, y"=(x+C)"=1.

Let's set an arbitrary initial condition y|_(x=x_0)=y_0 . Putting x=x_0 and y=y_0 in the equation y=x+C , we find that C=y_0-x_0 . Substituting this value of C into this function, we will have y=x+y_0-x_0 . This function satisfies the given initial condition: putting x=x_0 , we get y=x_0+y_0-x_0=y_0. So, the function y=x+C is the general solution of this equation.

In particular, setting x_0=0 and y_0=0 , we obtain a particular solution y=x .

The general solution of this equation, i.e. the function y=x+C , defines in the xOy plane a family of parallel lines with slope k=1 . A single integral line y=x+y_0-x_0 passes through each point M_0(x_0,y_0) of the xOy plane. The particular solution y=x determines one of the integral curves, namely the straight line passing through the origin (Fig. 4).

Example 2. Check that the function y=Ce^x is the general solution of the equation y"-y=0 and find a particular solution that satisfies the initial condition y|_(x=1)=-1. .

Solution. We have y=Ce^x,~y"=Ce^x . Substituting the expressions y and y" into this equation, we obtain Ce^x-Ce^x\equiv0 , i.e. the function y=Ce^x satisfies this equation for any values of the constant C .

Let's set an arbitrary initial condition y|_(x=x_0)=y_0 . Substituting x_0 and y_0 instead of x and y in the function y=Ce^x , we will have y_0=Ce^(x_0) , whence C=y_0e^(-x_0) . The function y=y_0e^(x-x_0) satisfies the initial condition. Indeed, assuming x=x_0 , we get y=y_0e^(x_0-x_0)=y_0. The function y=Ce^x is the general solution of this equation.

For x_0=1 and y_0=-1 we get a particular solution y=-e^(x-1) .

From a geometric point of view, the general solution defines a family of integral curves, which are the graphs of exponential functions; a particular solution is an integral curve passing through the point M_0(1;-1) (Fig.5).

A relation of the form \Phi(x,y,C)=0 , which implicitly determines the general solution, is called common integral differential equation of the first order.

The relation obtained from the general integral at a particular value of the constant C is called private integral differential equation.

The problem of solving or integrating a differential equation is to find a general solution or a general integral of a given differential equation. If an initial condition is additionally given, then it is required to select a particular solution or a particular integral that satisfies the given initial condition.

Since, from a geometric point of view, the x and y coordinates are equal, along with the equation \frac(dx)(dy)=f(x,y) we will consider the equation \frac(dx)(dy)=\frac(1)(f(x,y)).

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 called the process of finding solutions 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 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- not known features 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

Equations relating the independent variable, the desired function and its derivatives are called differential .

General form of differential equations: F (x, y, y’, y’’.. y’’’) = 0

Decision differential equation is a function that, when substituted into the equation, turns it into an identity.

The highest order of a derivative entering a DE is called in order this equation.

The process of finding a solution to the DE is called its integration .

First order differential equations

Ordinary differential equation of the first order is called an equation of the form F(x, y, y")=0, where F is a known function of three variables, x- independent variable, y(x) is the desired function, y"(x) is its derivative. If the equation F(x, y, y")=0 can be resolved relatively y", then it is written in the form y"=f(x, y)

The equation y"=f(x, y) establishes a relationship between the coordinates of the point ( x, y)and slopey" tangent to the integral curve passing through this point.

A first-order differential equation solved with respect to the derivative can be written as differential form :

P(x; y) dx+ Q(x; y) dy=0,

Where P(x; y) and Q(x; y) are known functions. The equation P(x; y) dx+ Q(x; y) dy=0 convenient because the variables in it are equal, i.e. any of them can be considered as a function of the other.

If the first order differential equation y"=f(x, y), has a solution, then, generally speaking, it has infinitely many solutions, and these solutions can be written in the form y=φ(x,C), where C is an arbitrary constant.

Functiony=φ(x,C) is called common solution differential equation of the 1st order. It contains one arbitrary constant and satisfies the following conditions:

Functiony=φ(x,C) is a solution to the DE for each fixed value FROM.

Whatever the initial condition y(x 0 )= y 0 , one can find such a value of the constant C=C 0 , what functiony=φ(x,C 0 ) satisfies the given initial condition.

Private decision DE of the first order is any function y=φ(x,C 0 ) obtained from the general solution y=φ(x,C) for a specific value of the constant C=C 0 .

The problem of finding a solution to the first-order DE P(x; y) dx+ Q(x; y) dy=0 , satisfying the given initial condition y(x 0 )= y 0 , is called Cauchy problem .

Theorem (existence and uniqueness of the solution of the Cauchy problem).

If in the equation y"=f(x, y) function f(x, y) and its partial derivative f" y (x, y) are continuous in some domain D, containing a dot (x 0 ; y 0 ), then there is only one solution y=φ(x)of this equation, satisfying the initial conditiony(x 0 )= y 0 . (no proof)

Separable Variable Equations

The simplest first-order DE is an equation of the form

P(x) dx+ Q(y) dy=0.

In it, one term depends only on x, and the other from y. Sometimes such DEs are called equations with separated variables . By integrating this equation term by term, we obtain:

∫ P(x) dx+ ∫ Q(y) dy= with - its general integral.

A more general case is described by equations with separable variables, which have the form:

P 1 (x) . Q 1 (y) . dx+P 2 (x) . Q 2 (y) . dy=0.

The peculiarity of this equation is that the coefficients are products of two functions, one of which depends only on X the other is only y.

The equation P 1 (x) . Q 1 (y) . dx+ P 2 (x) . Q 2 (y) . dy=0 easily reduced to the equation P(x) dx+ Q(y) dy=0. by term-by-term dividing it into Q 1 (y) . P 2 (x)≠0. We receive.

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 its solution is called, in which specific numerical values are assigned to arbitrary constants.

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 eat 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.

Educational Institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

FIRST ORDER DIFFERENTIAL EQUATIONS

Lecture summary for accounting students

correspondence form of education (NISPO)

Gorki, 2013

First order differential equations

The concept of a differential equation. General and particular solutions

When studying various phenomena, it is often not possible to find a law that directly connects the independent variable and the desired function, but it is possible to establish a connection between the desired function and its derivatives.

The relation connecting the independent variable, the desired function and its derivatives is called differential equation :

Here x is an independent variable, y is the desired function,
are the derivatives of the desired function. In this case, relation (1) requires the presence of at least one derivative.

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

Consider the differential equation

. (2)

Since this equation includes a derivative of only the first order, then it is called is a first-order differential equation.

If equation (2) can be solved with respect to the derivative and written as

, (3)

then such an equation is called a first-order differential equation in normal form.

In many cases it is expedient to consider an equation of the form

which is called a first-order differential equation written in differential form.

Because
, then equation (3) can be written as
or
, where one can count
and
. This means that equation (3) has been converted to equation (4).

We write equation (4) in the form
. Then
,
,
, where one can count
, i.e. an equation of the form (3) is obtained. Thus, equations (3) and (4) are equivalent.

By solving the differential equation (2) or (3) any function is called
, which, when substituting it into equation (2) or (3), turns it into an identity:

or
.

The process of finding all solutions of a differential equation is called its integration , and the solution graph
differential equation is called integral curve this equation.

If the solution of the differential equation is obtained in implicit form
, then it is called integral given differential equation.

General solution differential equation of the first order is a family of functions of the form
, depending on an arbitrary constant FROM, each of which is a solution of the given differential equation for any admissible value of an arbitrary constant FROM. Thus, the differential equation has an infinite number of solutions.

Private decision differential equation is called the solution obtained from the general solution formula for a specific value of an arbitrary constant FROM, including
.

The Cauchy problem and its geometric interpretation

Equation (2) has an infinite number of solutions. In order to single out one solution from this set, which is called a particular solution, some additional conditions must be specified.

The problem of finding a particular solution to equation (2) under given conditions is called Cauchy problem . This problem is one of the most important in the theory of differential equations.

The Cauchy problem is formulated as follows: among all solutions of equation (2) find such a solution
, in which the function
takes a given numeric value if the independent variable x takes a given numeric value , i.e.

,
, (5)

where D is the domain of the function
.

Meaning called the initial value of the function , a – initial value of the independent variable . Condition (5) is called initial condition or Cauchy condition .

From a geometric point of view, the Cauchy problem for differential equation (2) can be formulated as follows: from the set of integral curves of equation (2) select the one that passes through a given point
.

Differential equations with separable variables

One of the simplest types of differential equations is a first-order differential equation that does not contain the desired function:

. (6)

Given that
, we write the equation in the form
or
. Integrating both sides of the last equation, we get:
or

. (7)

Thus, (7) is a general solution to equation (6).

Example 1 . Find the general solution of the differential equation
.

Solution . We write the equation in the form
or
. We integrate both parts of the resulting equation:
,
. Let's finally write down
.

Example 2 . Find a solution to the equation
on condition
.

Solution . Let's find the general solution of the equation:
,
,
,
. By condition
,
. Substitute in the general solution:
or
. We substitute the found value of an arbitrary constant into the formula for the general solution:
. This is the particular solution of the differential equation that satisfies the given condition.

The equation

(8)

called a first-order differential equation that does not contain an independent variable . We write it in the form
or
. We integrate both parts of the last equation:
or
- general solution of equation (8).

Example . Find a general solution to the equation
.

Solution . We write this equation in the form:
or
. Then
,
,
,
. In this way,
is the general solution of this equation.

Type equation

(9)

integrated using separation of variables. To do this, we write the equation in the form
, and then, using the operations of multiplication and division, we bring it to such a form that one part includes only the function of X and differential dx, and in the second part - a function of at and differential dy. To do this, both sides of the equation must be multiplied by dx and divide by
. As a result, we obtain the equation

, (10)

in which the variables X and at separated. We integrate both parts of equation (10):
. The resulting relation is the general integral of equation (9).

Example 3 . Integrate Equation
.

Solution . Transform the equation and separate the variables:
,
. Let's integrate:
,
or is the general integral of this equation.
.

Let the equation be given in the form

Such an equation is called first-order differential equation with separable variables in symmetrical form.

To separate the variables, both sides of the equation must be divided by
:

. (12)

The resulting equation is called separated differential equation . We integrate equation (12):

.(13)

Relation (13) is a general integral of differential equation (11).

Example 4 . Integrate the differential equation.

Solution . We write the equation in the form

and divide both parts into
,
. The resulting equation:
is a separated variable equation. Let's integrate it:

,
,

,
. The last equality is the general integral of the given differential equation.

Example 5 . Find a particular solution of a differential equation
, satisfying the condition
.

Solution . Given that
, we write the equation in the form
or
. Let's separate the variables:
. Let's integrate this equation:
,
,
. The resulting relation is the general integral of this equation. By condition
. Substitute into the general integral and find FROM:
,FROM=1. Then the expression
is a particular solution of the given differential equation, written as a particular integral.

Linear differential equations of the first order

The equation

(14)

called linear differential equation of the first order . unknown function
and its derivative enter this equation linearly, and the functions
and
continuous.

If a
, then the equation

(15)

called linear homogeneous . If a
, then equation (14) is called linear inhomogeneous .

To find a solution to equation (14), one usually uses substitution method (Bernoulli) , the essence of which is as follows.

The solution of equation (14) will be sought in the form of a product of two functions

, (16)

where
and
- some continuous functions. Substitute
and derivative
into equation (14):

Function v will be chosen in such a way that the condition
. Then
. Thus, to find a solution to equation (14), it is necessary to solve the system of differential equations

The first equation of the system is a linear homogeneous equation and can be solved by the method of separation of variables:
,
,
,
,
. As a function
one can take one of the particular solutions of the homogeneous equation, i.e. at FROM=1:
. Substitute into the second equation of the system:
or
.Then
. Thus, the general solution of a first-order linear differential equation has the form
.

Example 6 . solve the equation
.

Solution . We will seek the solution of the equation in the form
. Then
. Substitute into the equation:

or
. Function v choose in such a way that the equality
. Then
. We solve the first of these equations by the method of separation of variables:
,
,
,
,. Function v Substitute into the second equation:
,
,
,
. The general solution to this equation is
.