Solution of linear homogeneous differential equations with constant coefficients. Second Order Linear Differential Equations with Constant Coefficients

The equation

where and are continuous functions in the interval is called an inhomogeneous second-order linear differential equation, the functions and are its coefficients. If in this interval, then the equation takes the form:

and is called a second-order homogeneous linear differential equation. If equation (**) has the same coefficients and as equation (*), then it is called a homogeneous equation corresponding to a non-homogeneous equation (*).

Homogeneous second-order linear differential equations

Let in the linear equation

And are constant real numbers.

We will seek a particular solution of the equation in the form of a function , where is a real or complex number to be determined. Differentiating with respect to , we get:

Substituting into the original differential equation, we get:

Hence, taking into account that , we have:

This equation is called the characteristic equation of a homogeneous linear differential equation. The characteristic equation also makes it possible to find . This is a second degree equation, so it has two roots. Let's denote them by and . Three cases are possible:

1) The roots are real and different. In this case common decision equations:

Example 1

2) The roots are real and equal. In this case, the general solution to the equation is:

Example2

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The characteristic equation has the form:

Solution of the characteristic equation:

General solution of the original differential equation:

3) Complex roots. In this case, the general solution to the equation is:

Example 3

The characteristic equation has the form:

Solution of the characteristic equation:

General solution of the original differential equation:

Inhomogeneous second-order linear differential equations

Consider now the solution of some types of linear not homogeneous equation second order with constant coefficients

where and are constant real numbers, is a known continuous function in the interval . To find the general solution of such a differential equation, it is necessary to know the general solution of the corresponding homogeneous differential equation and the particular solution. Let's consider some cases:

We are also looking for a particular solution of the differential equation in the form of a square trinomial:

If 0 is a single root of the characteristic equation, then

If 0 is a double root of the characteristic equation, then

The situation is similar if is a polynomial of arbitrary degree

Example 4

We solve the corresponding homogeneous equation.

Characteristic equation:

The general solution of the homogeneous equation:

Let us find a particular solution of the inhomogeneous dif-equation:

Substituting the found derivatives into the original differential equation, we obtain:

The desired particular solution:

General solution of the original differential equation:

We seek a particular solution in the form , where is an indeterminate coefficient.

Substituting and into the original differential equation, we obtain an identity, from which we find the coefficient.

If is the root of the characteristic equation, then we look for a particular solution of the original differential equation in the form , when is a single root, and , when is a double root.

Example 5

Characteristic equation:

The general solution of the corresponding homogeneous differential equation is:

Let us find a particular solution of the corresponding inhomogeneous differential equation:

The general solution of the differential equation:

In this case, we are looking for a particular solution in the form of a trigonometric binomial:

where and are uncertain coefficients

Substituting and into the original differential equation, we obtain an identity, from which we find the coefficients.

These equations determine the coefficients and except for the case when (or when are the roots of the characteristic equation). In the latter case, we look for a particular solution of the differential equation in the form:

Example6

Characteristic equation:

The general solution of the corresponding homogeneous differential equation is:

Let us find a particular solution of the inhomogeneous dif-equation

Substituting into the original differential equation, we get:

General solution of the original differential equation:

Number series convergence
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This article reveals the question of solving linear inhomogeneous second-order differential equations with constant coefficients. The theory will be considered along with examples of the given problems. For decryption incomprehensible terms it is necessary to refer to the topic of the basic definitions and concepts of the theory of differential equations.

Consider a linear differential equation (LDE) of the second order with constant coefficients of the form y "" + p y " + q y \u003d f (x) , where p and q are arbitrary numbers, and the existing function f (x) is continuous on the integration interval x .

Let us pass to the formulation of the general solution theorem for LIDE.

Yandex.RTB R-A-339285-1

General solution theorem for LDNU

Theorem 1

The general solution, located on the interval x, of an inhomogeneous differential equation of the form y (n) + f n - 1 (x) · y (n - 1) + . . . + f 0 (x) y = f (x) with continuous integration coefficients on x interval f 0 (x) , f 1 (x) , . . . , f n - 1 (x) and continuous function f (x) is equal to the sum of the general solution y 0 , which corresponds to the LODE and some particular solution y ~ , where the original inhomogeneous equation is y = y 0 + y ~ .

This shows that the solution of such a second-order equation has the form y = y 0 + y ~ . The algorithm for finding y 0 is considered in the article on linear homogeneous differential equations of the second order with constant coefficients. After that, one should proceed to the definition of y ~ .

The choice of a particular solution to the LIDE depends on the type of the available function f (x) located on the right side of the equation. To do this, it is necessary to consider separately the solutions of linear inhomogeneous differential equations of the second order with constant coefficients.

When f (x) is considered to be a polynomial of the nth degree f (x) = P n (x) , it follows that a particular solution of the LIDE is found by a formula of the form y ~ = Q n (x) x γ , where Q n ( x) is a polynomial of degree n, r is the number of zero roots of the characteristic equation. The value of y ~ is a particular solution y ~ "" + p y ~ " + q y ~ = f (x) , then the available coefficients, which are defined by the polynomial
Q n (x) , we find using the method of indefinite coefficients from the equality y ~ "" + p · y ~ " + q · y ~ = f (x) .

Example 1

Calculate using the Cauchy theorem y "" - 2 y " = x 2 + 1 , y (0) = 2 , y " (0) = 1 4 .

Solution

In other words, it is necessary to pass to a particular solution of a linear inhomogeneous differential equation of the second order with constant coefficients y "" - 2 y " = x 2 + 1 , which will satisfy the given conditions y (0) = 2 , y " (0) = 1 4 .

The general solution of a linear inhomogeneous equation is the sum of the general solution that corresponds to the equation y 0 or a particular solution of the inhomogeneous equation y ~ , that is, y = y 0 + y ~ .

First, let's find a general solution for the LNDE, and then a particular one.

Let's move on to finding y 0 . Writing the characteristic equation will help find the roots. We get that

k 2 - 2 k \u003d 0 k (k - 2) \u003d 0 k 1 \u003d 0, k 2 \u003d 2

We found that the roots are different and real. Therefore, we write

y 0 \u003d C 1 e 0 x + C 2 e 2 x \u003d C 1 + C 2 e 2 x.

Let's find y ~ . It's clear that right part given equation is a second degree polynomial, then one of the roots is equal to zero. From here we get that a particular solution for y ~ will be

y ~ = Q 2 (x) x γ \u003d (A x 2 + B x + C) x \u003d A x 3 + B x 2 + C x, where the values \u200b\u200bof A, B, C take undefined coefficients.

Let's find them from an equality of the form y ~ "" - 2 y ~ " = x 2 + 1 .

Then we get that:

y ~ "" - 2 y ~ " = x 2 + 1 (A x 3 + B x 2 + C x) "" - 2 (A x 3 + B x 2 + C x) " = x 2 + 1 3 A x 2 + 2 B x + C " - 6 A x 2 - 4 B x - 2 C = x 2 + 1 6 A x + 2 B - 6 A x 2 - 4 B x - 2 C = x 2 + 1 - 6 A x 2 + x (6 A - 4 B) + 2 B - 2 C = x 2 + 1

Equating the coefficients with the same exponents x , we get a system of linear expressions - 6 A = 1 6 A - 4 B = 0 2 B - 2 C = 1 . When solving in any of the ways, we find the coefficients and write: A \u003d - 1 6, B \u003d - 1 4, C \u003d - 3 4 and y ~ \u003d A x 3 + B x 2 + C x \u003d - 1 6 x 3 - 1 4 x 2 - 3 4 x .

This entry is called the general solution of the original linear inhomogeneous second-order differential equation with constant coefficients.

To find a particular solution that satisfies the conditions y (0) = 2 , y " (0) = 1 4 , it is required to determine the values C1 and C2, based on an equality of the form y \u003d C 1 + C 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x.

We get that:

y (0) = C 1 + C 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x x = 0 = C 1 + C 2 y "(0) = C 1 + C 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x "x = 0 = = 2 C 2 e 2 x - 1 2 x 2 + 1 2 x + 3 4 x = 0 = 2 C 2 - 3 4

We work with the resulting system of equations of the form C 1 + C 2 = 2 2 C 2 - 3 4 = 1 4 , where C 1 = 3 2 , C 2 = 1 2 .

Applying the Cauchy theorem, we have that

y = C 1 + C 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x = = 3 2 + 1 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x

Answer: 3 2 + 1 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x .

When the function f (x) is represented as a product of a polynomial with degree n and an exponent f (x) = P n (x) e a x , then from here we obtain that a particular solution of the second-order LIDE will be an equation of the form y ~ = e a x Q n ( x) · x γ , where Q n (x) is a polynomial of the nth degree, and r is the number of roots of the characteristic equation equal to α .

The coefficients belonging to Q n (x) are found by the equality y ~ "" + p · y ~ " + q · y ~ = f (x) .

Example 2

Find the general solution of a differential equation of the form y "" - 2 y " = (x 2 + 1) · e x .

Solution

The equation general view y = y 0 + y ~ . Specified Equation corresponds to the LOD y "" - 2 y " = 0. The previous example shows that its roots are equal to k1 = 0 and k 2 = 2 and y 0 = C 1 + C 2 e 2 x according to the characteristic equation.

It's clear that right side of the equation is x 2 + 1 · e x . From here, LNDE is found through y ~ = e a x Q n (x) x γ , where Q n (x) , which is a polynomial of the second degree, where α = 1 and r = 0 , because the characteristic equation does not have a root equal to 1 . Hence we get that

y ~ = e a x Q n (x) x γ = e x A x 2 + B x + C x 0 = e x A x 2 + B x + C .

A, B, C are unknown coefficients, which can be found by the equality y ~ "" - 2 y ~ " = (x 2 + 1) · e x .

Got that

y ~ "= e x A x 2 + B x + C" = e x A x 2 + B x + C + e x 2 A x + B == e x A x 2 + x 2 A + B + B + C y ~ "" = e x A x 2 + x 2 A + B + B + C " = = e x A x 2 + x 2 A + B + B + C + e x 2 A x + 2 A + B = = e x A x 2 + x 4 A + B + 2 A + 2 B + C

y ~ "" - 2 y ~ " = (x 2 + 1) e x ⇔ e x A x 2 + x 4 A + B + 2 A + 2 B + C - - 2 e x A x 2 + x 2 A + B + B + C = x 2 + 1 e x ⇔ e x - A x 2 - B x + 2 A - C = (x 2 + 1) e x ⇔ - A x 2 - B x + 2 A - C = x 2 + 1 ⇔ - A x 2 - B x + 2 A - C = 1 x 2 + 0 x + 1

We equate the indicators for the same coefficients and obtain a system of linear equations. From here we find A, B, C:

A = 1 - B = 0 2 A - C = 1 ⇔ A = - 1 B = 0 C = - 3

Answer: it can be seen that y ~ = e x (A x 2 + B x + C) = e x - x 2 + 0 x - 3 = - e x x 2 + 3 is a particular solution of LIDE, and y = y 0 + y = C 1 e 2 x - e x · x 2 + 3

When the function is written as f (x) = A 1 cos (β x) + B 1 sin β x , and A 1 and IN 1 are numbers, then an equation of the form y ~ = A cos β x + B sin β x x γ , where A and B are considered to be indefinite coefficients, and r the number of complex conjugate roots related to the characteristic equation, equal to ± i β . In this case, the search for coefficients is carried out by the equality y ~ "" + p · y ~ " + q · y ~ = f (x) .

Example 3

Find the general solution of a differential equation of the form y "" + 4 y = cos (2 x) + 3 sin (2 x) .

Solution

Before writing the characteristic equation, we find y 0 . Then

k 2 + 4 \u003d 0 k 2 \u003d - 4 k 1 \u003d 2 i, k 2 \u003d - 2 i

We have a pair of complex conjugate roots. Let's transform and get:

y 0 \u003d e 0 (C 1 cos (2 x) + C 2 sin (2 x)) \u003d C 1 cos 2 x + C 2 sin (2 x)

The roots from the characteristic equation are considered to be a conjugate pair ± 2 i , then f (x) = cos (2 x) + 3 sin (2 x) . This shows that the search for y ~ will be made from y ~ = (A cos (β x) + B sin (β x) x γ = (A cos (2 x) + B sin (2 x)) x. Unknowns coefficients A and B will be sought from an equality of the form y ~ "" + 4 y ~ = cos (2 x) + 3 sin (2 x) .

Let's transform:

y ~ " = ((A cos (2 x) + B sin (2 x) x) " = = (- 2 A sin (2 x) + 2 B cos (2 x)) x + A cos (2 x) + B sin (2 x) y ~ "" = ((- 2 A sin (2 x) + 2 B cos (2 x)) x + A cos (2 x) + B sin (2 x)) " = = (- 4 A cos (2 x) - 4 B sin (2 x)) x - 2 A sin (2 x) + 2 B cos (2 x) - - 2 A sin (2 x) + 2 B cos (2 x) = = (- 4 A cos (2 x) - 4 B sin (2 x)) x - 4 A sin (2 x) + 4 B cos (2 x)

Then it is seen that

y ~ "" + 4 y ~ = cos (2 x) + 3 sin (2 x) ⇔ (- 4 A cos (2 x) - 4 B sin (2 x)) x - 4 A sin (2 x) + 4 B cos (2 x) + + 4 (A cos (2 x) + B sin (2 x)) x = cos (2 x) + 3 sin (2 x) ⇔ - 4 A sin (2 x) + 4B cos(2x) = cos(2x) + 3 sin(2x)

It is necessary to equate the coefficients of sines and cosines. We get a system of the form:

4 A = 3 4 B = 1 ⇔ A = - 3 4 B = 1 4

It follows that y ~ = (A cos (2 x) + B sin (2 x) x = - 3 4 cos (2 x) + 1 4 sin (2 x) x .

Answer: the general solution of the original LIDE of the second order with constant coefficients is considered to be

y = y 0 + y ~ = = C 1 cos (2 x) + C 2 sin (2 x) + - 3 4 cos (2 x) + 1 4 sin (2 x) x

When f (x) = e a x P n (x) sin (β x) + Q k (x) cos (β x) , then y ~ = e a x (L m (x) sin (β x) + N m (x) cos (β x) x γ We have that r is the number of complex conjugate pairs of roots related to the characteristic equation, equal to α ± i β , where P n (x) , Q k (x) , L m (x) and N m (x) are polynomials of degree n, k, m, where m = m a x (n, k). Finding coefficients L m (x) and N m (x) is produced based on the equality y ~ "" + p · y ~ " + q · y ~ = f (x) .

Example 4

Find the general solution y "" + 3 y " + 2 y = - e 3 x ((38 x + 45) sin (5 x) + (8 x - 5) cos (5 x)) .

Solution

It is clear from the condition that

α = 3 , β = 5 , P n (x) = - 38 x - 45 , Q k (x) = - 8 x + 5 , n = 1 , k = 1

Then m = m a x (n , k) = 1 . We find y 0 by first writing the characteristic equation of the form:

k 2 - 3 k + 2 = 0 D = 3 2 - 4 1 2 = 1 k 1 = 3 - 1 2 = 1, k 2 = 3 + 1 2 = 2

We found that the roots are real and distinct. Hence y 0 = C 1 e x + C 2 e 2 x . Next, it is necessary to look for a general solution based on an inhomogeneous equation y ~ of the form

y ~ = e α x (L m (x) sin (β x) + N m (x) cos (β x) x γ = = e 3 x ((A x + B) cos (5 x) + (C x + D) sin (5 x)) x 0 = = e 3 x ((A x + B) cos (5 x) + (C x + D) sin (5 x))

It is known that A, B, C are coefficients, r = 0, because there is no pair of conjugate roots related to the characteristic equation with α ± i β = 3 ± 5 · i . These coefficients are found from the resulting equality:

y ~ "" - 3 y ~ " + 2 y ~ = - e 3 x ((38 x + 45) sin (5 x) + (8 x - 5) cos (5 x)) ⇔ (e 3 x (( A x + B) cos (5 x) + (C x + D) sin (5 x))) "" - - 3 (e 3 x ((A x + B) cos (5 x) + (C x + D) sin (5 x))) = - e 3 x ((38 x + 45) sin (5 x) + (8 x - 5) cos (5 x))

Finding the derivative and similar terms gives

E 3 x ((15 A + 23 C) x sin (5 x) + + (10 A + 15 B - 3 C + 23 D) sin (5 x) + + (23 A - 15 C) x cos (5 x) + (- 3 A + 23 B - 10 C - 15 D) cos (5 x)) = = - e 3 x (38 x sin (5 x) + 45 sin (5 x) + + 8 x cos (5 x) - 5 cos (5 x))

After equating the coefficients, we obtain a system of the form

15 A + 23 C = 38 10 A + 15 B - 3 C + 23 D = 45 23 A - 15 C = 8 - 3 A + 23 B - 10 C - 15 D = - 5 ⇔ A = 1 B = 1 C = 1 D = 1

From all it follows that

y ~= e 3 x ((A x + B) cos (5 x) + (C x + D) sin (5 x)) == e 3 x ((x + 1) cos (5 x) + (x+1)sin(5x))

Answer: now the general solution of the given linear equation:

y = y 0 + y ~ = = C 1 e x + C 2 e 2 x + e 3 x ((x + 1) cos (5 x) + (x + 1) sin (5 x))

Algorithm for solving LDNU

Definition 1

Any other kind of function f (x) for the solution provides for the solution algorithm:

finding the general solution of the corresponding linear homogeneous equation, where y 0 = C 1 ⋅ y 1 + C 2 ⋅ y 2 , where y 1 and y2 are linearly independent particular solutions of LODE, From 1 and From 2 are considered arbitrary constants;
acceptance as a general solution of the LIDE y = C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2 ;
definition of derivatives of a function through a system of the form C 1 "(x) + y 1 (x) + C 2 "(x) y 2 (x) = 0 C 1 "(x) + y 1" (x) + C 2 " (x) y 2 "(x) = f (x) , and finding functions C 1 (x) and C 2 (x) through integration.

Example 5

Find the general solution for y "" + 36 y = 24 sin (6 x) - 12 cos (6 x) + 36 e 6 x .

Solution

We proceed to writing the characteristic equation, having previously written y 0 , y "" + 36 y = 0 . Let's write and solve:

k 2 + 36 = 0 k 1 = 6 i , k 2 = - 6 i ⇒ y 0 = C 1 cos (6 x) + C 2 sin (6 x) ⇒ y 1 (x) = cos (6 x) , y 2 (x) = sin (6 x)

We have that the record of the general solution of the given equation will take the form y = C 1 (x) cos (6 x) + C 2 (x) sin (6 x) . It is necessary to pass to the definition of derivative functions C 1 (x) and C2(x) according to the system with equations:

C 1 "(x) cos (6 x) + C 2" (x) sin (6 x) = 0 C 1 "(x) (cos (6 x))" + C 2 "(x) (sin (6 x)) " = 0 ⇔ C 1 " (x) cos (6 x) + C 2 " (x) sin (6 x) = 0 C 1 " (x) (- 6 sin (6 x) + C 2 "(x) (6 cos (6 x)) \u003d \u003d 24 sin (6 x) - 12 cos (6 x) + 36 e 6 x

A decision needs to be made regarding C 1 "(x) and C2" (x) using any method. Then we write:

C 1 "(x) \u003d - 4 sin 2 (6 x) + 2 sin (6 x) cos (6 x) - 6 e 6 x sin (6 x) C 2 "(x) \u003d 4 sin (6 x) cos (6 x) - 2 cos 2 (6 x) + 6 e 6 x cos (6 x)

Each of the equations must be integrated. Then we write the resulting equations:

C 1 (x) = 1 3 sin (6 x) cos (6 x) - 2 x - 1 6 cos 2 (6 x) + + 1 2 e 6 x cos (6 x) - 1 2 e 6 x sin ( 6 x) + C 3 C 2 (x) = - 1 6 sin (6 x) cos (6 x) - x - 1 3 cos 2 (6 x) + + 1 2 e 6 x cos (6 x) + 1 2 e 6 x sin (6 x) + C 4

It follows that the general solution will have the form:

y = 1 3 sin (6 x) cos (6 x) - 2 x - 1 6 cos 2 (6 x) + + 1 2 e 6 x cos (6 x) - 1 2 e 6 x sin (6 x) + C 3 cos (6 x) + + - 1 6 sin (6 x) cos (6 x) - x - 1 3 cos 2 (6 x) + + 1 2 e 6 x cos (6 x) + 1 2 e 6 x sin (6 x) + C 4 sin (6 x) = = - 2 x cos (6 x) - x sin (6 x) - 1 6 cos (6 x) + + 1 2 e 6 x + C 3 cos (6 x) + C 4 sin (6 x)

Answer: y = y 0 + y ~ = - 2 x cos (6 x) - x sin (6 x) - 1 6 cos (6 x) + + 1 2 e 6 x + C 3 cos (6 x) + C 4 sin (6x)

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Here we apply the method of variation of the Lagrange constants to solve linear inhomogeneous second-order differential equations. Detailed description this method for solving equations of arbitrary order is set out on the page
Solution of linear inhomogeneous differential equations of higher orders by the Lagrange method >>> .

Example 1

Solve a second-order differential equation with constant coefficients using the variation of Lagrange constants:
(1)

Solution

First, we solve the homogeneous differential equation:
(2)

This is a second order equation.

We solve the quadratic equation:
.
Multiple roots: . Fundamental system solutions of equation (2) has the form:
(3) .
Hence we obtain the general solution of the homogeneous equation (2):
(4) .

We vary the constants C 1 and C 2 . That is, we replace the constants and in (4) with functions:
.
We are looking for a solution to the original equation (1) in the form:
(5) .

We find the derivative:
.
We connect the functions and the equation:
(6) .
Then
.

We find the second derivative:
.
We substitute into the original equation (1):
(1) ;

.
Since and satisfy the homogeneous equation (2), the sum of the terms in each column of the last three rows is zero, and the previous equation becomes:
(7) .
Here .

Together with equation (6), we obtain a system of equations for determining the functions and :
(6) :
(7) .

Solving a system of equations

We solve the system of equations (6-7). Let's write expressions for functions and :
.
We find their derivatives:
;
.

We solve the system of equations (6-7) by the Cramer method. We calculate the determinant of the matrix of the system:

.
By Cramer's formulas we find:
;
.

So, we found derivatives of functions:
;
.
Let's integrate (see Methods of integrating roots). Making a substitution
; ; ; .

.
.

;
.

Answer

Example 2

Solve the differential equation by the method of variation of the Lagrange constants:
(8)

Solution

Step 1. Solution of the homogeneous equation

We solve a homogeneous differential equation:

(9)
Looking for a solution in the form . We compose the characteristic equation:

This equation has complex roots:
.
The fundamental system of solutions corresponding to these roots has the form:
(10) .
The general solution of the homogeneous equation (9):
(11) .

Step 2. Variation of Constants - Replacing Constants with Functions

Now we vary the constants C 1 and C 2 . That is, we replace the constants in (11) with functions:
.
We are looking for a solution to the original equation (8) in the form:
(12) .

Further, the course of the solution is the same as in example 1. We arrive at the following system of equations for determining the functions and :
(13) :
(14) .
Here .

Solving a system of equations

Let's solve this system. Let's write out the expressions of the functions and :
.
From the table of derivatives we find:
;
.

We solve the system of equations (13-14) by the Cramer method. System matrix determinant:

.
By Cramer's formulas we find:
;
.

.
Since , then the modulus sign under the logarithm sign can be omitted. Multiply the numerator and denominator by:
.
Then
.

General solution of the original equation:

.

Linear homogeneous differential equation of the second order with constant coefficients has a general solution
, where and linearly independent particular solutions of this equation.

General form of solutions of a second-order homogeneous differential equation with constant coefficients
, depends on the roots of the characteristic equation
.

The roots of the characteristic equations	Kind of general solution
Roots and valid and various
Roots == valid and identical
Complex roots ,

Example

Find the general solution of linear homogeneous differential equations of the second order with constant coefficients:

Solution:
.

Having solved it, we will find the roots
,
valid and different. Therefore, the general solution is:
.

Solution: Let's make the characteristic equation:
.

Having solved it, we will find the roots

valid and identical. Therefore, the general solution is:
.

Solution: Let's make the characteristic equation:
.

Having solved it, we will find the roots
complex. Therefore, the general solution is:

Linear inhomogeneous second-order differential equation with constant coefficients has the form

Where
. (1)

The general solution of a linear inhomogeneous second-order differential equation has the form
, where
is a particular solution of this equation, is a general solution of the corresponding homogeneous equation, i.e. equations.

Type of private decision
inhomogeneous equation (1) depending on the right side
:

Right part	Private decision
– degree polynomial	, where is the number of roots of the characteristic equation equal to zero.
	, where = is the root of the characteristic equation.
	Where - number, equal to the number roots of the characteristic equation coinciding with .
	where is the number of roots of the characteristic equation coinciding with .

Consider different types of right-hand sides of a linear non-homogeneous differential equation:

1.
, where is a polynomial of degree . Then a particular solution
can be searched in the form
, where

, a is the number of roots of the characteristic equation equal to zero.

Example

Find a general solution
.

Solution:

B) Since the right side of the equation is a polynomial of the first degree and none of the roots of the characteristic equation
not equal to zero (
), then we look for a particular solution in the form where and are unknown coefficients. Differentiating twice
and substituting
,
and
into the original equation, we find.

Equating the coefficients at the same powers on both sides of the equation
,
, we find
,
. So, a particular solution of this equation has the form
, and its general solution.

2. Let the right side look like
, where is a polynomial of degree . Then a particular solution
can be searched in the form
, where
is a polynomial of the same degree as
, a - a number indicating how many times is the root of the characteristic equation.

Example

Find a general solution
.

Solution:

A) Find the general solution of the corresponding homogeneous equation
. To do this, we write the characteristic equation
. Let's find the roots of the last equation
. Therefore, the general solution of the homogeneous equation has the form
.

characteristic equation

, where is an unknown coefficient. Differentiating twice
and substituting
,
and
into the original equation, we find. Where
, that is
or
.

So, a particular solution of this equation has the form
, and its general solution
.

3. Let the right side look like , where
and - given numbers. Then a particular solution
can be searched in the form where and are unknown coefficients, and is a number equal to the number of roots of the characteristic equation coinciding with
. If in a function expression
include at least one of the functions
or
, then in
should always be entered both functions.

Example

Find a general solution .

Solution:

B) Since the right side of the equation is a function
, then the control number of this equation, it does not coincide with the roots
characteristic equation
. Then we look for a particular solution in the form

Where and are unknown coefficients. Differentiating twice, we get. Substituting
,
and
into the original equation, we find

Bringing like terms together, we get

We equate the coefficients at
and
on the right and left sides of the equation, respectively. We get the system
. Solving it, we find
,
.

So, a particular solution of the original differential equation has the form .

The general solution of the original differential equation has the form .

Educational Institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Guidelines

on the study of the topic "Linear differential equations of the second order" by students of the accounting department of the correspondence form of education (NISPO)

Gorki, 2013

Linear differential equations

second order with constantcoefficients

Linear homogeneous differential equations

Linear differential equation of the second order with constant coefficients is called an equation of the form

those. an equation that contains the desired function and its derivatives only to the first degree and does not contain their products. In this equation and
are some numbers, and the function
given on some interval
.

If a
on the interval
, then equation (1) takes the form

, (2)

and called linear homogeneous . Otherwise, equation (1) is called linear inhomogeneous .

Consider the complex function

, (3)

where
and
are real functions. If function (3) is a complex solution of equation (2), then the real part
, and the imaginary part
solutions
taken separately are solutions of the same homogeneous equation. Thus, any complex solution of equation (2) generates two real solutions of this equation.

Solutions of a homogeneous linear equation have the following properties:

If a is a solution to equation (2), then the function
, where FROM- an arbitrary constant, will also be a solution to equation (2);

If a and are solutions of equation (2), then the function
will also be a solution to equation (2);

If a and are solutions of equation (2), then their linear combination
will also be a solution to equation (2), where and
are arbitrary constants.

Functions
and
called linearly dependent on the interval
if there are such numbers and
, not zero at the same time that on this interval the equality

If equality (4) holds only when
and
, then the functions
and
called linearly independent on the interval
.

Example 1 . Functions
and
are linearly dependent, since
along the whole number line. In this example
.

Example 2 . Functions
and
are linearly independent on any interval, since the equality
possible only if and
, and
.

Construction of a general solution of a linear homogeneous

equations

In order to find a general solution to equation (2), you need to find two of its linearly independent solutions and . Linear combination of these solutions
, where and
are arbitrary constants, and will give the general solution of a linear homogeneous equation.

Linearly independent solutions of Eq. (2) will be sought in the form

, (5)

where - some number. Then
,
. Let us substitute these expressions into equation (2):

or
.

Because
, then
. So the function
will be a solution to equation (2) if will satisfy the equation

. (6)

Equation (6) is called characteristic equation for equation (2). This equation is an algebraic quadratic equation.

Let and are the roots of this equation. They can be either real and different, or complex, or real and equal. Let's consider these cases.

Let the roots and characteristic equations are real and distinct. Then the solutions of equation (2) will be the functions
and
. These solutions are linearly independent, since the equality
can only be performed when
, and
. Therefore, the general solution of Eq. (2) has the form

where and
are arbitrary constants.

Example 3
.

Solution . The characteristic equation for this differential will be
. Solving it quadratic equation, find its roots
and
. Functions
and
are solutions of the differential equation. The general solution of this equation has the form
.

complex number is called an expression of the form
, where and are real numbers, and
is called the imaginary unit. If a
, then the number
is called purely imaginary. If
, then the number
is identified with a real number .

Number is called the real part of the complex number, and - the imaginary part. If two complex numbers differ from each other only in the sign of the imaginary part, then they are called conjugate:
,
.

Example 4 . Solve a quadratic equation
.

Solution . Equation discriminant
. Then. Likewise,
. Thus, this quadratic equation has conjugate complex roots.

Let the roots of the characteristic equation be complex, i.e.
,
, where
. Solutions to equation (2) can be written as
,
or
,
. According to Euler's formulas

,
.

Then ,. As is known, if a complex function is a solution of a linear homogeneous equation, then the solutions of this equation are both the real and imaginary parts of this function. Thus, the solutions of equation (2) will be the functions
and
. Since equality

can only be performed if
and
, then these solutions are linearly independent. Therefore, the general solution of equation (2) has the form

where and
are arbitrary constants.

Example 5 . Find the general solution of the differential equation
.

Solution . The equation
is characteristic for the given differential. We solve it and get complex roots
,
. Functions
and
are linearly independent solutions of the differential equation. The general solution of this equation has the form.

Let the roots of the characteristic equation be real and equal, i.e.
. Then the solutions of equation (2) are the functions
and
. These solutions are linearly independent, since the expression can be identically equal to zero only when
and
. Therefore, the general solution of equation (2) has the form
.

Example 6 . Find the general solution of the differential equation
.

Solution . Characteristic equation
has equal roots
. In this case, the linearly independent solutions of the differential equation are the functions
and
. The general solution has the form
.

Inhomogeneous second-order linear differential equations with constant coefficients

and special right side

The general solution of the linear inhomogeneous equation (1) is equal to the sum of the general solution
corresponding homogeneous equation and any particular solution
inhomogeneous equation:
.

In some cases, a particular solution of an inhomogeneous equation can be found quite simply by the form of the right side
equations (1). Let's consider cases when it is possible.

those. the right side of the inhomogeneous equation is a polynomial of degree m. If a
is not a root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form of a polynomial of degree m, i.e.

Odds
are determined in the process of finding a particular solution.

If
is the root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form

Example 7 . Find the general solution of the differential equation
.

Solution . The corresponding homogeneous equation for this equation is
. Its characteristic equation
has roots
and
. The general solution of the homogeneous equation has the form
.

Because
is not a root of the characteristic equation, then we will seek a particular solution of the inhomogeneous equation in the form of a function
. Find the derivatives of this function
,
and substitute them into this equation:

or . Equate the coefficients at and free members:
Deciding this system, we get
,
. Then a particular solution of the inhomogeneous equation has the form
, and the general solution of this inhomogeneous equation will be the sum of the general solution of the corresponding homogeneous equation and the particular solution of the inhomogeneous one:
.

Let the inhomogeneous equation have the form

If a
is not a root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form. If
is the root of the characteristic multiplicity equation k (k=1 or k=2), then in this case the particular solution of the inhomogeneous equation will have the form .

Example 8 . Find the general solution of the differential equation
.

Solution . The characteristic equation for the corresponding homogeneous equation has the form
. its roots
,
. In this case, the general solution of the corresponding homogeneous equation is written as
.

Since the number 3 is not the root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form
. Let's find derivatives of the first and second orders:,

Substitute into the differential equation:
+ +,
+,.

Equate the coefficients at and free members:

From here
,
. Then a particular solution of this equation has the form
, and the general solution

Lagrange method of variation of arbitrary constants

The method of variation of arbitrary constants can be applied to any inhomogeneous linear equation with constant coefficients, regardless of the form of the right side. This method makes it possible to always find a general solution to an inhomogeneous equation if the general solution of the corresponding homogeneous equation is known.

Let
and
are linearly independent solutions of Eq. (2). Then the general solution to this equation is
, where and
are arbitrary constants. The essence of the method of variation of arbitrary constants is that the general solution of equation (1) is sought in the form

where
and
- new unknown features to be found. Since there are two unknown functions, two equations containing these functions are needed to find them. These two equations make up the system

which is a linear algebraic system of equations with respect to
and
. Solving this system, we find
and
. Integrating both parts of the obtained equalities, we find

and
.

Substituting these expressions into (9), we obtain the general solution of the inhomogeneous linear equation (1).

Example 9 . Find the general solution of the differential equation
.

Solution. The characteristic equation for the homogeneous equation corresponding to the given differential equation is
. Its roots are complex
,
. Because
and
, then
,
, and the general solution of the homogeneous equation has the form Then the general solution of this inhomogeneous equation will be sought in the form where
and
- unknown functions.

The system of equations for finding these unknown functions has the form

Solving this system, we find
,
. Then

,
. Let us substitute the obtained expressions into the general solution formula:

This is the general solution of this differential equation obtained by the Lagrange method.

Questions for self-control of knowledge

Which differential equation is called a second-order linear differential equation with constant coefficients?

Which linear differential equation is called homogeneous, and which one is called non-homogeneous?

What are the properties of a linear homogeneous equation?

What equation is called characteristic for a linear differential equation and how is it obtained?

In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of different roots of the characteristic equation?

In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of equal roots of the characteristic equation?

In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of complex roots of the characteristic equation?

How is the general solution of a linear inhomogeneous equation written?

In what form is a particular solution of a linear inhomogeneous equation sought if the roots of the characteristic equation are different and not equal to zero, and the right side of the equation is a polynomial of degree m?

In what form is a particular solution of a linear inhomogeneous equation sought if there is one zero among the roots of the characteristic equation, and the right side of the equation is a polynomial of degree m?

What is the essence of the Lagrange method?