Method of variation of arbitrary constants. Examples of the method of variation of an arbitrary constant

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 the 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) .

External links

• exponenta.ru - Theoretical information with examples

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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 1st order linear equation.

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

Consider a linear inhomogeneous differential equation of the first order:
(1) .
There are three ways to solve this equation:

• method of variation of constant (Lagrange).

Let's consider solving a first-order linear differential equation using the Lagrange method.

Method of variation of constant (Lagrange)

In the variation of constant method, we solve the equation in two steps. In the first step, we simplify the original equation and solve a homogeneous equation. At the second stage, we replace the constant of integration obtained at the first stage of the solution with a function. Then we look for a general solution to the original equation.

Consider the equation:
(1)

Step 1 Solving a homogeneous equation

We are looking for a solution to the homogeneous equation:

This is a separable equation

We separate the variables - multiply by dx, divide by y:

Let's integrate:

Integral over y - tabular:

Then

Let's potentiate:

Let's replace the constant e C with C and remove the modulus sign, which comes down to multiplying by a constant ±1, which we will include in C:

Step 2 Replace the constant C with the function

Now let's replace the constant C with a function of x:
C → u (x)
That is, we will look for a solution to the original equation (1) as:
(2)
Finding the derivative.

According to the rule of differentiation of a complex function:
.
According to the product differentiation rule:

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

.
Two members are reduced:
;
.
Let's integrate:
.
Substitute in (2) :
.
As a result, we obtain a general solution to a first-order linear differential equation:
.

An example of solving a first-order linear differential equation by the Lagrange method

Solve the equation

Solution

We solve the homogeneous equation:

We separate the variables:

Multiply by:

Let's integrate:

Tabular integrals:

Let's potentiate:

Let's replace the constant e C with C and remove the modulus signs:

From here:

Let's replace the constant C with a function of x:
C → u (x)

Finding the derivative:
.
Substitute into the original equation:
;
;
Or:
;
.
Let's integrate:
;
Solution of the equation:
.