Find a solution to the equation using the method of varying arbitrary constants. Method of variation of arbitrary constants

Let us now consider the linear inhomogeneous equation
. (2)
Let y 1 ,y 2 ,.., y n - fundamental system decisions, and - common decision appropriate homogeneous equation L(y)=0 . Similar to the case of first-order equations, we will look for a solution to equation (2) in the form
. (3)
Let us make sure that a solution in this form exists. To do this, we substitute the function into the equation. To substitute this function into the equation, we find its derivatives. The first derivative is equal to
. (4)
When calculating the second derivative, four terms will appear on the right side of (4), when calculating the third derivative, eight terms will appear, and so on. Therefore, for the convenience of further calculations, the first term in (4) is set equal to zero. Taking this into account, the second derivative is equal to
. (5)
For the same reasons as before, in (5) we also set the first term equal to zero. Finally, nth derivative equal to
. (6)
Substituting the obtained values of the derivatives into the original equation, we have
. (7)
The second term in (7) is equal to zero, since the functions y j , j=1,2,..,n, are solutions to the corresponding homogeneous equation L(y)=0. Combining with the previous one, we get the system algebraic equations to find functions C" j (x)
(8)
The determinant of this system is the Wronski determinant of the fundamental system of solutions y 1 ,y 2 ,..,y n of the corresponding homogeneous equation L(y)=0 and therefore is not equal to zero. Consequently, there is a unique solution to system (8). Having found it, we obtain the functions C" j (x), j=1,2,…,n, and, consequently, C j (x), j=1,2,…,n Substituting these values into (3), we obtain a solution to a linear inhomogeneous equation.
The presented method is called the method of variation of an arbitrary constant or the Lagrange method.

Example No. 1. Let's find the general solution of the equation y"" + 4y" + 3y = 9e -3 x. Consider the corresponding homogeneous equation y"" + 4y" + 3y = 0. Its roots characteristic equation r 2 + 4r + 3 = 0 are equal to -1 and -3. Therefore, the fundamental system of solutions to a homogeneous equation consists of the functions y 1 = e - x and y 2 = e -3 x. We look for a solution to the inhomogeneous equation in the form y = C 1 (x)e - x + C 2 (x)e -3 x. To find the derivatives C" 1 , C" 2 we compose a system of equations (8)

solving which, we find , Integrating the obtained functions, we have

Finally we get

Example No. 2. Solve linear differential equations second order with constant coefficients by the method of varying arbitrary constants:

y(0) =1 + 3ln3
y’(0) = 10ln3

Solution:
This differential equation refers to linear differential equations with constant coefficients.
We will look for a solution to the equation in the form y = e rx. To do this, we compose the characteristic equation of a linear homogeneous differential equation with constant coefficients:
r 2 -6 r + 8 = 0
D = (-6) 2 - 4 1 8 = 4

Roots of the characteristic equation: r 1 = 4, r 2 = 2
Consequently, the fundamental system of solutions consists of the functions:
y 1 = e 4x , y 2 = e 2x
The general solution of the homogeneous equation has the form:

Search for a particular solution by the method of varying an arbitrary constant.
To find the derivatives of C" i we compose a system of equations:

C" 1 (4e 4x) + C" 2 (2e 2x) = 4/(2+e -2x)
Let's express C" 1 from the first equation:
C" 1 = -c 2 e -2x
and substitute it into the second one. As a result we get:
C" 1 = 2/(e 2x +2e 4x)
C" 2 = -2e 2x /(e 2x +2e 4x)
We integrate the obtained functions C" i:
C 1 = 2ln(e -2x +2) - e -2x + C * 1
C 2 = ln(2e 2x +1) – 2x+ C * 2

Because the , then we write the resulting expressions in the form:
C 1 = (2ln(e -2x +2) - e -2x + C * 1) e 4x = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x
C 2 = (ln(2e 2x +1) – 2x+ C * 2)e 2x = e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
Thus, the general solution to the differential equation has the form:
y = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x + e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
or
y = 2 e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + C * 1 e 4x + C * 2 e 2x

Let's find a particular solution under the condition:
y(0) =1 + 3ln3
y’(0) = 10ln3

Substituting x = 0 into the found equation, we get:
y(0) = 2 ln(3) - 1 + ln(3) + C * 1 + C * 2 = 3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
We find the first derivative of the obtained general solution:
y’ = 2e 2x (2C 1 e 2x + C 2 -2x +4 e 2x ln(e -2x +2)+ ln(2e 2x +1)-2)
Substituting x = 0, we get:
y’(0) = 2(2C 1 + C 2 +4 ln(3)+ ln(3)-2) = 4C 1 + 2C 2 +10 ln(3) -4 = 10ln3

We get a system of two equations:
3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
4C 1 + 2C 2 +10 ln(3) -4 = 10ln3
or
C*1+C*2=2
4C 1 + 2C 2 = 4
or
C*1+C*2=2
2C 1 + C 2 = 2
Where:
C 1 = 0, C * 2 = 2
The private solution will be written as:
y = 2 e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + 2 e 2x

Theoretical minimum
In the theory of differential equations, there is a method that claims to have a fairly high degree of universality for this theory.
We are talking about the method of variation of an arbitrary constant, applicable to solving various classes of differential equations and their
systems This is precisely the case when the theory - if we take the proofs of the statements out of brackets - is minimal, but allows us to achieve
significant results, so the emphasis will be on examples.
The general idea of the method is quite simple to formulate. Let given equation(system of equations) is difficult to solve or completely incomprehensible,
how to solve it. However, it is clear that by eliminating some terms from the equation, it is solved. Then they solve exactly this simplified
equation (system), we obtain a solution containing a certain number of arbitrary constants - depending on the order of the equation (the number
equations in the system). Then it is assumed that the constants in the found solution are not actually constants; the found solution
is substituted into the original equation (system), a differential equation (or system of equations) is obtained to determine the “constants”.
There is a certain specificity in the application of the method of variation of an arbitrary constant to different tasks, but these are already particulars that will be
demonstrated with examples.
Let us separately consider the solution of linear inhomogeneous equations higher orders, i.e. equations of the form
.
The general solution of a linear inhomogeneous equation is the sum of the general solution of the corresponding homogeneous equation and a particular solution
given equation. Let us assume that a general solution to the homogeneous equation has already been found, namely, a fundamental system of solutions (FSS) has been constructed
. Then the general solution of the homogeneous equation is equal to .
We need to find any particular solution to the inhomogeneous equation. For this purpose, constants are considered to depend on a variable.
Next you need to solve the system of equations
.
The theory guarantees that this system of algebraic equations with respect to derivatives of functions has a unique solution.
When finding the functions themselves, the constants of integration do not appear: after all, any one solution is sought.
In the case of solving systems of linear inhomogeneous first-order equations of the form

the algorithm remains almost unchanged. First you need to find the FSR of the corresponding homogeneous system of equations, compose the fundamental matrix
system, the columns of which represent the elements of the FSR. Next, the equation is drawn up
.
When solving the system, we determine the functions , thus finding a particular solution to the original system
(the fundamental matrix is multiplied by the column of found functions).
We add it to the general solution of the corresponding system of homogeneous equations, which is constructed on the basis of the already found FSR.
The general solution of the original system is obtained.
Examples.
Example 1. Linear inhomogeneous equations of the first order.
Let us consider the corresponding homogeneous equation (we denote the desired function):
.
This equation can easily be solved using the separation of variables method:

.
Now let’s imagine the solution to the original equation in the form , where the function has yet to be found.
We substitute this type of solution into the original equation:
.
As you can see, the second and third terms on the left side cancel each other out - this is characteristic method of variation of an arbitrary constant.

Here it is already a truly arbitrary constant. Thus,
.
Example 2. Bernoulli's equation.
We proceed similarly to the first example - we solve the equation

method of separation of variables. It turns out, so we look for a solution to the original equation in the form
.
We substitute this function into the original equation:
.
And again the reductions occur:
.
Here you need to remember to make sure that when dividing by the solution is not lost. And the solution to the original one corresponds to the case
equations Let's remember it. So,
.
Let's write it down.
This is the solution. When writing the answer, you should also indicate the previously found solution, since it does not correspond to any final value
constants
Example 3. Linear inhomogeneous equations of higher orders.
Let us immediately note that this equation can be solved more simply, but it is convenient to demonstrate the method using it. Although some advantages
The variation method has an arbitrary constant in this example too.
So, you need to start with the FSR of the corresponding homogeneous equation. Let us recall that to find the FSR, a characteristic curve is compiled
the equation
.
Thus, the general solution of the homogeneous equation
.
The constants included here must be varied. Making up a system

The method of variation of an arbitrary constant, or the Lagrange method, is another way to solve first-order linear differential equations and the Bernoulli equation.

Linear differential equations of the first order are equations of the form y’+p(x)y=q(x). If there is a zero on the right side: y’+p(x)y=0, then this is a linear homogeneous 1st order equation. Accordingly, an equation with nonzero right side, y’+p(x)y=q(x), — heterogeneous linear equation 1st order.

Method of variation of an arbitrary constant (Lagrange method) is as follows:

1) We are looking for a general solution to the homogeneous equation y’+p(x)y=0: y=y*.

2) In the general solution, we consider C not a constant, but a function of x: C = C (x). We find the derivative of the general solution (y*)’ and substitute the resulting expression for y* and (y*)’ into the initial condition. From the resulting equation we find the function C(x).

3) In the general solution of the homogeneous equation, instead of C, we substitute the found expression C(x).

Let's look at examples of the method of varying an arbitrary constant. Let's take the same tasks as in, compare the progress of the solution and make sure that the answers obtained coincide.

1) y’=3x-y/x

Let's rewrite the equation in standard form (unlike Bernoulli's method, where we needed the notation form only to see that the equation is linear).

y’+y/x=3x (I). Now we proceed according to plan.

1) Solve the homogeneous equation y’+y/x=0. This is an equation with separable variables. Imagine y’=dy/dx, substitute: dy/dx+y/x=0, dy/dx=-y/x. We multiply both sides of the equation by dx and divide by xy≠0: dy/y=-dx/x. Let's integrate:

2) In the resulting general solution of the homogeneous equation, we will consider C not a constant, but a function of x: C=C(x). From here

We substitute the resulting expressions into condition (I):

Let's integrate both sides of the equation:

here C is already some new constant.

3) In the general solution of the homogeneous equation y=C/x, where we assumed C=C(x), that is, y=C(x)/x, instead of C(x) we substitute the found expression x³+C: y=(x³ +C)/x or y=x²+C/x. We got the same answer as when solving by Bernoulli's method.

Answer: y=x²+C/x.

2) y’+y=cosx.

Here the equation is already written in standard form; there is no need to transform it.

1) Solve the homogeneous linear equation y’+y=0: dy/dx=-y; dy/y=-dx. Let's integrate:

To obtain a more convenient form of notation, we take the exponent to the power of C as the new C:

This transformation was performed to make it more convenient to find the derivative.

2) In the resulting general solution of the linear homogeneous equation, we consider C not a constant, but a function of x: C=C(x). Under this condition

We substitute the resulting expressions y and y’ into the condition:

Multiply both sides of the equation by

We integrate both sides of the equation using the integration by parts formula, we get:

Here C is no longer a function, but an ordinary constant.

3) In the general solution of the homogeneous equation

substitute the found function C(x):

We got the same answer as when solving by Bernoulli's method.

The method of variation of an arbitrary constant is also applicable to solve.

y'x+y=-xy².

We reduce the equation to standard view: y’+y/x=-y² (II).

1) Solve the homogeneous equation y’+y/x=0. dy/dx=-y/x. We multiply both sides of the equation by dx and divide by y: dy/y=-dx/x. Now let's integrate:

We substitute the resulting expressions into condition (II):

Let's simplify:

We obtained an equation with separable variables for C and x:

Here C is already an ordinary constant. During the integration process, we wrote simply C instead of C(x), so as not to overload the notation. And at the end we returned to C(x), so as not to confuse C(x) with the new C.

3) In the general solution of the homogeneous equation y=C(x)/x we substitute the found function C(x):

We got the same answer as when solving it using the Bernoulli method.

Self-test examples:

1. Let's rewrite the equation in standard form: y’-2y=x.

1) Solve the homogeneous equation y’-2y=0. y’=dy/dx, hence dy/dx=2y, multiply both sides of the equation by dx, divide by y and integrate:

From here we find y:

We substitute the expressions for y and y’ into the condition (for brevity we will use C instead of C(x) and C’ instead of C"(x)):

To find the integral on the right side, we use the integration by parts formula:

Now we substitute u, du and v into the formula:

Here C =const.

3) Now we substitute homogeneous into the solution

Consider a linear inhomogeneous differential equation with constant coefficients of arbitrary nth order:
(1) .
The method of variation of a constant, which we considered for a first-order equation, is also applicable for higher-order equations.

The solution is carried out in two stages. In the first step, we discard the right-hand side and solve the homogeneous equation. As a result, we obtain a solution containing n arbitrary constants. At the second stage we vary the constants. That is, we believe that these constants are functions of the independent variable x and find the form of these functions.

Although we are considering equations with constant coefficients here, but Lagrange's method is also applicable to solving any linear inhomogeneous equations. To do this, however, the fundamental system of solutions to the homogeneous equation must be known.

Step 1. Solving the homogeneous equation

As in the case of first-order equations, we first look for the general solution of the homogeneous equation, equating the right-hand inhomogeneous side to zero:
(2) .
The general solution to this equation is:
(3) .
Here are arbitrary constants; - n linearly independent solutions of homogeneous equation (2), which form a fundamental system of solutions to this equation.

Step 2. Variation of constants - replacing constants with functions

In the second stage we will deal with the variation of constants. In other words, we will replace the constants with functions of the independent variable x:
.
That is, we are looking for a solution to the original equation (1) in the following form:
(4) .

If we substitute (4) into (1), we get one differential equation for n functions. In this case, we can connect these functions with additional equations. Then you get n equations from which n functions can be determined. Additional equations can be written different ways. But we will do this so that the solution has the simplest form. To do this, when differentiating, you need to equate to zero the terms containing derivatives of the functions. Let's demonstrate this.

To substitute the proposed solution (4) into the original equation (1), we need to find the derivatives of the first n orders of the function written in the form (4). We differentiate (4) using rules for differentiating sums and works:
.
Let's group the members. First, we write down the terms with derivatives of , and then the terms with derivatives of :

.
Let's impose the first condition on the functions:
(5.1) .
Then the expression for the first derivative with respect to will have a simpler form:
(6.1) .

Using the same method, we find the second derivative:

.
Let's impose a second condition on the functions:
(5.2) .
Then
(6.2) .
And so on. In additional conditions, we equate terms containing derivatives of functions to zero.

Thus, if we choose the following additional equations for the functions:
(5.k) ,
then the first derivatives with respect to will have the simplest form:
(6.k) .
Here .

Find the nth derivative:
(6.n)
.

Substitute into the original equation (1):
(1) ;

.
Let us take into account that all functions satisfy equation (2):
.
Then the sum of terms containing zero gives zero. As a result we get:
(7) .

As a result, we received a system of linear equations for derivatives:
(5.1) ;
(5.2) ;
(5.3) ;
. . . . . . .
(5.n-1) ;
(7′) .

Solving this system, we find expressions for derivatives as a function of x. Integrating, we get:
.
Here are constants that no longer depend on x. Substituting into (4), we obtain a general solution to the original equation.

Note that to determine the values of the derivatives, we never used the fact that the coefficients a i are constant. That's why Lagrange's method is applicable to solve any linear inhomogeneous equations, if the fundamental system of solutions to the homogeneous equation (2) is known.

Examples

Solve equations using the method of variation of constants (Lagrange).

Let us turn to the consideration of linear inhomogeneous differential equations of the form

Where - the required function of the argument , and the functions

are given and continuous on a certain interval
.

Let us introduce into consideration a linear homogeneous equation, the left side of which coincides with the left side of the inhomogeneous equation (2.31),

An equation of the form (2.32) is called homogeneous equation corresponding to the inhomogeneous equation (2.31).

The following theorem holds about the structure of the general solution of the inhomogeneous linear equation (2.31).

Theorem 2.6. The general solution of the linear inhomogeneous equation (2.31) in the region

is the sum of any particular solution of it and the general solution of the corresponding homogeneous equation (2.32) in the domain (2.33), i.e.

Where - particular solution of equation (2.31),
is the fundamental system of solutions to the homogeneous equation (2.32), and
- arbitrary constants.

You will find the proof of this theorem in.

Using the example of a second-order differential equation, we will outline a method by which one can find a particular solution to a linear inhomogeneous equation. This method is called Lagrange method of variation of arbitrary constants.

So, let us be given an inhomogeneous linear equation

(2.35)

where are the coefficients
and right side
continuous in some interval
.

Let us denote by
And
fundamental system of solutions to the homogeneous equation

(2.36)

Then its general solution has the form

(2.37)

Where And - arbitrary constants.

We will look for a solution to equation (2.35) in the same form , as well as the general solution of the corresponding homogeneous equation, replacing arbitrary constants with some differentiable functions of (we vary arbitrary constants), those.

Where
And
- some differentiable functions from , which are still unknown and which we will try to determine so that function (2.38) would be a solution to the inhomogeneous equation (2.35). Differentiating both sides of equality (2.38), we obtain

So that when calculating second order derivatives of
And
, we require that everywhere in
the condition was met

Then for will have

Let's calculate the second derivative

Substituting expressions for ,,from (2.38), (2.40), (2.41) into equation (2.35), we obtain

Expressions in square brackets, are equal to zero everywhere in
, because And - partial solutions of equation (2.36). In this case, (2.42) will take the form Combining this condition with condition (2.39), we obtain a system of equations for determining
And

(2.43)

The last system is a system of two algebraic linear inhomogeneous equations with respect to
And
. The determinant of this system is the Wronski determinant for the fundamental system of solutions ,and, therefore, is nonzero everywhere in
. This means that system (2.43) has a unique solution. Having solved it in any way relatively
,
we'll find

Where
And
- known functions.

Performing integration and taking into account that as
,
we should take one pair of functions and set the integration constants equal to zero. We get

Substituting expressions (2.44) into relations (2.38), we can write the desired solution to the inhomogeneous equation (2.35) in the form

This method can be generalized to find a particular solution to the linear inhomogeneous equation -th order.

Example 2.6. Solve the equation
at
if functions

form a fundamental system of solutions to the corresponding homogeneous equation.

Let's find a particular solution to this equation. To do this, in accordance with the Lagrange method, we must first solve system (2.43), which in our case has the form
Reducing both sides of each equation by we get

Subtracting the first equation term by term from the second equation, we find
and then from the first equation it follows
Performing integration and setting the integration constants to zero, we will have

A particular solution to this equation can be represented as

The general solution of this equation has the form

Where And - arbitrary constants.

Finally, let us note one remarkable property, which is often called the principle of superposition of solutions and is described by the following theorem.

Theorem 2.7. If in between
function
- particular solution of the equation function
a particular solution of the equation on the same interval is the function
there is a particular solution to the equation