Method of variation of constants for linear ones. Solution of linear inhomogeneous differential equations of higher orders by the Lagrangian method. Method of variation of arbitrary constants for constructing a solution to a linear inhomogeneous differential equation

Let us now consider the linear inhomogeneous equation
. (2)
Let y 1 ,y 2 ,.., y n be a fundamental system of solutions, and let be the general solution of the corresponding 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, the nth derivative is
. (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 obtain a system of algebraic equations for finding the 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 to the equation y"" + 4y" + 3y = 9e -3 x. Consider the corresponding homogeneous equation y"" + 4y" + 3y = 0. The roots of its 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)
C′ 1 ·e -x +C′ 2 ·e -3x =0
-C′ 1 ·e -x -3C′ 2 ·e -3x =9e -3x
solving which, we find , Integrating the obtained functions, we have
Finally we get

Example No. 2. Solve second-order linear differential equations with constant coefficients using 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: y =C 1 e 4x +C 2 e 2x
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 ·e 4x +C′ 2 ·e 2x =0
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

Since y =C 1 ·e 4x +C 2 ·e 2x, 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
From: C 1 = 0, C * 2 = 2
The private solution will be written as:
y = 2e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + 2 e 2x

The method of variation of arbitrary constants is used to solve inhomogeneous differential equations. This lesson is intended for those students who are already more or less well versed in the topic. If you are just starting to get acquainted with remote control, i.e. If you are a teapot, I recommend starting with the first lesson: First order differential equations. Examples of solutions. And if you are already finishing, please discard the possible preconception that the method is difficult. Because it's simple.

In what cases is the method of variation of arbitrary constants used?

1) The method of variation of an arbitrary constant can be used to solve linear inhomogeneous DE of the 1st order. Since the equation is of the first order, then the constant is also one.

2) The method of variation of arbitrary constants is used to solve some linear inhomogeneous second order equations. Here two constants vary.

It is logical to assume that the lesson will consist of two paragraphs... So I wrote this sentence, and for about 10 minutes I was painfully thinking about what other clever crap I could add for a smooth transition to practical examples. But for some reason I don’t have any thoughts after the holidays, although I don’t seem to have abused anything. Therefore, let’s get straight to the first paragraph.

Method of variation of an arbitrary constant
for a first order linear inhomogeneous equation

Before considering the method of variation of an arbitrary constant, it is advisable to be familiar with the article Linear differential equations of the first order. In that lesson we practiced first solution inhomogeneous 1st order DE. This first solution, I remind you, is called replacement method or Bernoulli method(not to be confused with Bernoulli's equation!!!)

Now we will look second solution– method of variation of an arbitrary constant. I will give only three examples, and I will take them from the above-mentioned lesson. Why so few? Because in fact, the solution in the second way will be very similar to the solution in the first way. In addition, according to my observations, the method of variation of arbitrary constants is used less frequently than the replacement method.

Example 1

(Diffur from Example No. 2 of the lesson Linear inhomogeneous differential equations of the 1st order)

Solution: This equation is linear inhomogeneous and has a familiar form:

At the first stage, it is necessary to solve a simpler equation:
That is, we stupidly reset the right side and write zero instead.
The equation I'll call auxiliary equation.

In this example, you need to solve the following auxiliary equation:

Before us separable equation, the solution of which (I hope) is no longer difficult for you:

Thus:
– general solution of the auxiliary equation.

On the second step we will replace some constant for now unknown function that depends on "x":

Hence the name of the method - we vary the constant. Alternatively, the constant could be some function that we now have to find.

IN original inhomogeneous equation let's make a replacement:

Let's substitute and into the equation :

Control point – the two terms on the left side cancel. If this does not happen, you should look for the error above.

As a result of the replacement, an equation with separable variables was obtained. We separate the variables and integrate.

What a blessing, the exponents also cancel:

We add a “normal” constant to the found function:

On final stage Let's remember our replacement:

The function has just been found!

So the general solution is:

Answer: common decision:

If you print out the two solutions, you will easily notice that in both cases we found the same integrals. The only difference is in the solution algorithm.

Now for something more complicated, I will also comment on the second example:

Example 2

Find the general solution to the differential equation
(Diffur from Example No. 8 of lesson Linear inhomogeneous differential equations of the 1st order)

Solution: Let us reduce the equation to the form :

Let's reset the right-hand side and solve the auxiliary equation:

General solution to the auxiliary equation:

In the inhomogeneous equation we make the replacement:

According to the product differentiation rule:

Let's substitute and into the original inhomogeneous equation:

The two terms on the left side cancel, which means we are on the right track:

Let's integrate by parts. The tasty letter from the integration by parts formula is already involved in the solution, so we use, for example, the letters “a” and “be”:

Now let's remember the replacement:

Answer: common decision:

And one example for independent decision:

Example 3

Find a particular solution to the differential equation corresponding to the given initial condition.

,
(Diffur from Example No. 4 of the lesson Linear inhomogeneous differential equations of the 1st order)
Solution:
This DE is linear inhomogeneous. We use the method of variation of arbitrary constants. Let's solve the auxiliary equation:

We separate the variables and integrate:

Common decision:
In the inhomogeneous equation we make the replacement:

Let's perform the substitution:

So the general solution is:

Let us find a particular solution corresponding to the given initial condition:

Answer: private solution:

The solution at the end of the lesson can serve as an example for finishing the assignment.

Method of variation of arbitrary constants
for a linear inhomogeneous second order equation
with constant coefficients

I have often heard the opinion that the method of varying arbitrary constants for a second-order equation is not an easy thing. But I assume the following: most likely, the method seems difficult to many because it does not occur so often. But in reality there are no particular difficulties - the course of the decision is clear, transparent, and understandable. And beautiful.

To master the method, it is desirable to be able to solve inhomogeneous second-order equations by selecting a particular solution based on the form of the right-hand side. This method is discussed in detail in the article. Inhomogeneous 2nd order DEs. We recall that a second-order linear inhomogeneous equation with constant coefficients has the form:

The selection method, which was discussed in the above lesson, works only in a limited number of cases when the right side contains polynomials, exponentials, sines, and cosines. But what to do when on the right, for example, is a fraction, logarithm, tangent? In such a situation, the method of variation of constants comes to the rescue.

Example 4

Find the general solution to a second order differential equation

Solution: On the right side given equation there is a fraction, so we can immediately say that the method of selecting a particular solution does not work. We use the method of variation of arbitrary constants.

There are no signs of a thunderstorm; the beginning of the solution is completely ordinary:

We'll find common decision appropriate homogeneous equations:

Let's compose and solve the characteristic equation:


– conjugate complex roots are obtained, so the general solution is:

Pay attention to the record of the general solution - if there are parentheses, then open them.

Now we do almost the same trick as for the first-order equation: we vary the constants, replacing them with unknown functions. That is, general solution of inhomogeneous we will look for equations in the form:

Where - for now unknown functions.

It looks like a household waste dump, but now we'll sort everything out.

The unknowns are the derivatives of the functions. Our goal is to find derivatives, and the found derivatives must satisfy both the first and second equations of the system.

Where do the “Greeks” come from? The stork brings them. We look at the general solution obtained earlier and write:

Let's find the derivatives:

The left parts have been dealt with. What's on the right?

is the right side of the original equation, in in this case:

The coefficient is the coefficient of the second derivative:

In practice, almost always, and our example is no exception.

Everything is clear, now you can create a system:

The system is usually solved according to Cramer's formulas using the standard algorithm. The only difference is that instead of numbers we have functions.

Let's find the main determinant of the system:

If you have forgotten how the two-by-two determinant is revealed, refer to the lesson How to calculate the determinant? The link leads to the board of shame =)

So: this means that the system has a unique solution.

Finding the derivative:

But that's not all, so far we have only found the derivative.
The function itself is restored by integration:

Let's look at the second function:

Here we add a “normal” constant

At the final stage of the solution, we remember in what form we were looking for a general solution to the inhomogeneous equation? In such:

The functions you need have just been found!

All that remains is to perform the substitution and write down the answer:

Answer: common decision:

In principle, the answer could have expanded the parentheses.

A complete check of the answer is carried out according to the standard scheme, which was discussed in the lesson. Inhomogeneous 2nd order DEs. But the verification will not be easy, since it is necessary to find rather heavy derivatives and carry out cumbersome substitution. This is an unpleasant feature when you solve such diffusers.

Example 5

Solve a differential equation by varying arbitrary constants

This is an example for you to solve on your own. In fact, on the right side there is also a fraction. Let's remember trigonometric formula, by the way, it will need to be applied during the solution.

The method of variation of arbitrary constants is the most universal method. It can solve any equation that can be solved method of selecting a particular solution based on the form of the right-hand side. The question arises: why not use the method of variation of arbitrary constants there too? The answer is obvious: the selection of a particular solution, which was discussed in class Inhomogeneous second order equations, significantly speeds up the solution and shortens the recording - no fuss with determinants and integrals.

Let's look at two examples with Cauchy problem.

Example 6

Find a particular solution to the differential equation corresponding to the given initial conditions

,

Solution: Again the fraction and exponent in interesting place.
We use the method of variation of arbitrary constants.

We'll find common decision appropriate homogeneous equations:



– different real roots are obtained, so the general solution is:

General solution of inhomogeneous we look for equations in the form: , where – for now unknown functions.

Let's create a system:

In this case:
,
Finding derivatives:
,

Thus:

Let's solve the system using Cramer's formulas:
, which means the system has a unique solution.

We restore the function by integration:

Used here method of subsuming a function under the differential sign.

We restore the second function by integration:

This integral is solved variable replacement method:

From the replacement itself we express:

Thus:

This integral can be found complete square extraction method, but in examples with diffusers I prefer to expand the fraction method of undetermined coefficients:

Both functions found:

As a result, the general solution to the inhomogeneous equation is:

Let's find a particular solution that satisfies the initial conditions .

Technically, the search for a solution is carried out in a standard way, which was discussed in the article Inhomogeneous differential equations of the second order.

Hold on, now we will find the derivative of the found general solution:

This is such a disgrace. It is not necessary to simplify it; it is easier to immediately create a system of equations. According to the initial conditions :

Let's substitute the found values ​​of the constants to the general solution:

In the answer, the logarithms can be packed a little.

Answer: private solution:

As you can see, difficulties may arise in integrals and derivatives, but not in the algorithm itself for the method of variation of arbitrary constants. It’s not me who intimidated you, it’s all Kuznetsov’s collection!

For relaxation, a final, simpler example for solving it yourself:

Example 7

Solve the Cauchy problem

,

The example is simple, but creative, when you create a system, look at it carefully before deciding ;-),




As a result, the general solution is:

Let us find a particular solution corresponding to the initial conditions .



Let us substitute the found values ​​of the constants into the general solution:

Answer: private solution:

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 applying the method of variation of an arbitrary constant to different problems, but these are already specifics that will
demonstrated with examples.

Let us separately consider the solution of linear inhomogeneous equations of 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
of this 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 .
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 a non-zero right-hand side, y’+p(x)y=q(x), is 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 received 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 received the same answer as when solving by Bernoulli's method.

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

y'x+y=-xy².

We bring the equation to standard form: 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

Method of variation of arbitrary constants

Method of variation of arbitrary constants for constructing a solution to a linear inhomogeneous differential equation

a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = f(t)

consists of replacing arbitrary constants c k in the general solution

z(t) = c 1 z 1 (t) + c 2 z 2 (t) + ... + c n z n (t)

corresponding homogeneous equation

a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = 0

for auxiliary functions c k (t) , whose derivatives satisfy the linear algebraic system

The determinant of system (1) is the Wronskian of the functions z 1 ,z 2 ,...,z n , which ensures its unique solvability with respect to .

If are antiderivatives for , taken at fixed values ​​of the integration constants, then the function

is a solution to the original linear inhomogeneous differential equation. Integration of an inhomogeneous equation in the presence of a general solution to the corresponding homogeneous equation is thus reduced to quadratures.

Method of variation of arbitrary constants for constructing solutions to a system of linear differential equations in vector normal form

consists in constructing a particular solution (1) in the form

Where Z(t) is the basis of solutions to the corresponding homogeneous equation, written in the form of a matrix, and vector function, which replaced the vector of arbitrary constants, is defined by the relation . The required particular solution (with zero initial values ​​at t = t 0 looks like

For a system with constant coefficients, the last expression is simplified:

Matrix Z(t)Z− 1 (τ) called Cauchy matrix operator L = A(t) .