The solution of linear homogeneous differential equations with constant coefficients. Linear second-order differential equations with constant coefficients

The equation

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

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

Uniform differential linear equations of the second order

Suppose in the linear equation

And - constant valid numbers.

The particular solution of the equation will be signed as a function where is a valid or complex number to be determined. Differentiating by, we get:

Substituting in the original diferration, we get:

From here, considering that we have:

This equation is called the characteristic equation of homogeneous linear difraration. Characteristic equation and makes it possible to find. This is the second equation, therefore has two roots. Denote them through and. Three cases are possible:

1) The roots are valid and different. In this case common decision Equations:

Example 1.

2) Roots are valid and equal. In this case, the general solution of the equation:

Example2

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The characteristic equation is:

Solution of the characteristic equation:

General solution of initial difraration:

3) Roots are integrated. In this case, the general solution of the equation:

Example 3.

The characteristic equation is:

Solution of the characteristic equation:

General solution of initial difraration:

Inhomogeneous differential linear equations of the second order

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

where and are constant valid numbers, a well-known continuous function in the interval. To find a general solution of such a differential equation, it is necessary to know the overall solution of the corresponding homogeneous differential equation and the particular solution. Consider some cases:

The particular solution of the differential equation is also looking for in the form of square three declections:

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

If 0 is a two-time root of a characteristic equation, then

The situation is similar if it is a polynomial of random

Example 4.

Resolving the appropriate homogeneous equation.

Characteristic equation:

General solution of a homogeneous equation:

We find a private solution of inhomogeneous difraration:

Substituting the found derivatives in the initial diferration, we get:

Second private solution:

General solution of initial difraration:

Private solution we are looking for in the form where - an indefinite coefficient.

Substituting and in the original differential equation, we obtain identity, from where we find the coefficient.

If - the root of the characteristic equation, then the private solution of the initial differential equation is looking for in the form when is a single root, and when there is a two-time root.

Example 5.

Characteristic equation:

General solution of the corresponding homogeneous differential equation:

We find a private solution of the corresponding inhomogeneous differential equation:

General Development Solution:

In this case, a particular solution is looking for in the form of trigonometric bicno:

where and - uncertain coefficients

Substituting and in the original differential equation, we will receive identity, from where we find the coefficients.

These equations determine the coefficients and except for the case when (or when the roots of the characteristic equation). In the latter case, a particular solution to the differential equation is looking for:

Example6

Characteristic equation:

General solution of the corresponding homogeneous difraration:

Find a private solution of inhomogeneous difraration

Substituting in the original diferration, we get:

General solution of initial difraration:

The convergence of numerical row
The definition of the convergence of the series and the tasks for the study of the convergence of numerical rows are considered in detail - signs of comparison, the sign of the convergence of Dalamber, the sign of Cauchy convergence and the integral sign of Cauchy convergence.

Absolute and conditional convergence of a series
The page discusses the alternating rows, their conditional and absolute convergence, the sign of the convergence of the leibher for the alignment of the rows - is contained. brief theory The subject and an example of solving the problem.

This article discloses the question of solving linear inhomogeneous second-order differential equations with permanent coefficients. The theory will be considered together with the examples of the given tasks. For decryption incomprehensible terms It is necessary to contact the topic of the basic definitions and concepts of the theory of differential equations.

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

Let us turn to the wording of the LFD general decision theorem.

Yandex.rtb R-A-339285-1

General decision theorem LDNU

Theorem 1.

A general solution that is at the interval of the inhomogeneous differential equation of the form y (n) + F n - 1 (x) · y (n - 1) +. . . + F 0 (x) · y \u003d F (x) with continuous integration coefficients on the x interval F 0 (x), F 1 (x) ,. . . , f n - 1 (x) and continuous function F (x) is equal to the sum of the overall solution y 0, which corresponds to a log and some particular solution y ~, where the initial inhomogeneous equation is y \u003d y 0 + y ~.

It can be seen that the solution of such an equation of the second order has the form y \u003d y 0 + y ~. The algorithm of finding Y 0 is considered in the article on linear homogeneous second-order differential equations with constant coefficients. After that, you should move to the definition Y ~.

The choice of the LFD private solution depends on the view of the existing function f (x), located in the right part of the equation. To do this, it is necessary to consider separately solutions of linear inhomogeneous second-order differential equations with constant coefficients.

When f (x) is considered for a polynomial N-degree f (x) \u003d p n (x), it follows that the LFD's private solution is found according to the formula y ~ \u003d Q n (x) · x γ, where Q n ( x) is a polynomial degree n, R is the number of zero roots of the characteristic equation. The value y ~ is a private solution Y ~ "" + p · y ~ "+ q · y ~ \u003d f (x), then the available coefficients that are defined by the polynomial
Q n (x), find it with the help of the method of uncertain coefficients from the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).

Example 1.

Calculate on the Cauchy theorem y "" - 2 y "\u003d x 2 + 1, y (0) \u003d 2, y" (0) \u003d 1 4.

Decision

In other words, it is necessary to move to a private solution of the linear inhomogeneous differential equation of the second order with constant coefficients Y "" - 2 y "\u003d x 2 + 1, which will satisfy the specified conditions y (0) \u003d 2, y" (0) \u003d 1 4 .

The general solution of the linear inhomogeneous equation is the sum of the overall solution, which corresponds to the equation y 0 or the private solution of the inhomogeneous Y ~ equation, that is, y \u003d y 0 + y ~.

To begin with, we will find a general solution for the LFD, and after which is private.

Let us turn to finding y 0. Recording a characteristic equation will help to 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

Received that the roots are different and valid. So write

y 0 \u003d C 1 E 0 X + C 2 E 2 X \u003d C 1 + C 2 E 2 x.

We find y ~. It's clear that right part specified equation It is a polynomial of a second degree, then one of the roots is zero. From here we get that a private solution for Y ~ will

y ~ \u003d 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\u200bA, B, C are taken by indefinite coefficients.

We find them from the equality of the form y ~ "" - 2 y ~ "\u003d x 2 + 1.

Then we get that:

y ~ "" - 2 y ~ "\u003d x 2 + 1 (a x 3 + b x 2 + c x)" "- 2 (a x 3 + b x 2 + c x)" \u003d x 2 + 1 3 a x 2 + 2 b x + c "- 6 a x 2 - 4 b x - 2 c \u003d x 2 + 1 6 a x + 2 b - 6 a x 2 - 4 b x - 2 c \u003d x 2 + 1 - 6 A x 2 + x (6 A - 4 B) + 2 B - 2 C \u003d x 2 + 1

Equating the coefficients with the same indicators of degrees X, we obtain a system of linear expressions - 6 a \u003d 1 6 A - 4 B \u003d 0 2 B - 2 C \u003d 1. When solving 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 differential equation of the second order with constant coefficients.

To find a private solution that satisfies the conditions Y (0) \u003d 2, Y "(0) \u003d 1 4, it is required to determine the values C 1 and C 2.Based on the 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) \u003d C 1 + C 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 xx \u003d 0 \u003d C 1 + C 2 y "(0) \u003d C 1 + C 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x "x \u003d 0 \u003d 2 c 2 e 2 x - 1 2 x 2 + 1 2 x + 3 4 x \u003d 0 \u003d 2 C 2 - 3 4

We work with the obtained system of equations of the form C 1 + C 2 \u003d 2 2 C 2 - 3 4 \u003d 1 4, where C 1 \u003d 3 2, C 2 \u003d 1 2.

Applying Cauchy theorem, we have that

y \u003d C 1 + C 2 E 2 X - 1 6 x 3 + 1 4 x 2 + 3 4 x \u003d 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 presented in the form of a piece of polynomial with a degree n and exponent F (x) \u003d p n (x) · EAX, then it turns out that the equation of the form y ~ \u003d EAX · Q n ( x) · X γ, where Q n (x) is a polynomial of n-degree, and R is the number of roots of the characteristic equation equal to α.

The coefficients belonging to Q n (x) are on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).

Example 2.

Find the general solution of the differential equation of the form y "" - 2 y "\u003d (x 2 + 1) · e x.

Decision

The equation general view y \u003d y 0 + y ~. Specified equation corresponds to the log y "" - 2 y "\u003d 0. According to the previous example, it can be seen that its roots are equal K 1 \u003d 0 and k 2 \u003d 2 and y 0 \u003d C 1 + C 2 E 2 x on a characteristic equation.

It's clear that the right part The equations are x 2 + 1 · e x. From here, the LFD is located through Y ~ \u003d E A x · Q n (x) · x γ, where q n (x), which is a polynomial of a second degree, where α \u003d 1 and r \u003d 0, because there is no root equal to 1. From here we get that

y ~ \u003d E A x · Q n (x) · x γ \u003d e x · a x 2 + b x + c · x 0 \u003d e x · a x 2 + b x + c.

A, B, C are unknown coefficients that can be found on the equality y ~ "" - 2 y ~ "\u003d (x 2 + 1) · e x.

Received that

y ~ "\u003d ex · a x 2 + b x + c" \u003d ex · a x 2 + b x + c + ex · 2 a x + b \u003d ex · a x 2 + x 2 a + b + b + C y ~ "" \u003d ex · a x 2 + x 2 a + b + b + c "\u003d \u003d ex · a x 2 + x 2 a + b + b + c + ex · 2 a x + 2 a + b \u003d \u003d EX · A x 2 + x 4 A + B + 2 A + 2 B + C

y ~ "" - 2 y ~ "\u003d (x 2 + 1) · ex ⇔ ex · a x 2 + x 4 a + b + 2 a + 2 b + c - - 2 ex · a x 2 + x 2 a + B + b + c \u003d x 2 + 1 · ex ⇔ ex · a x 2 - b x + 2 a - c \u003d (x 2 + 1) · ex ⇔ - a x 2 - b x + 2 A - C \u003d x 2 + 1 ⇔ - a x 2 - b x + 2 a - c \u003d 1 · x 2 + 0 · x + 1

Indicators with the same coefficients equate and obtain a system of linear equations. From here and find A, B, C:

A \u003d 1 - b \u003d 0 2 A - C \u003d 1 ⇔ A \u003d - 1 B \u003d 0 C \u003d - 3

Answer: It can be seen that y ~ \u003d ex · (a x 2 + b x + c) \u003d ex · - x 2 + 0 · x - 3 \u003d - ex · x 2 + 3 is a special solution of the LFD, and y \u003d y 0 + y \u003d C 1 E 2 x - Ex И x 2 + 3 is a general solution for inhomogeneous second-order dyferal.

When the function is written as f (x) \u003d a 1 cos (β x) + b 1 sin β x, and A 1. and IN 1are numbers, then the LDDa equation y ~ \u003d A cos β x + b sin β x · x γ is considered to be a private solution of the LLD, where A and B are considered uncertain coefficients, and R by the number of complex conjugate roots belonging to a characteristic equation equal to ± I β . In this case, the search for coefficients is carried out on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).

Example 3.

Find a general solution of the differential equation of the form y "" + 4 y \u003d cos (2 x) + 3 sin (2 x).

Decision

Before writing a 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 couple of comprehensively conjugate roots. We 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) \u003d cos (2 x) + 3 sin (2 x). It can be seen that the search y ~ will be made from Y ~ \u003d (a COS (β x) + b sin (β x) · x γ \u003d (A COS (2 x) + b sin (2 x)) · x. Unknown The coefficients A and B will be seen from the equality of the form y ~ "" + 4 y ~ \u003d cos (2 x) + 3 sin (2 x).

We transform:

y ~ "\u003d ((a COS (2 x) + b sin (2 x) · x)" \u003d \u003d (- 2 A sin (2 x) + 2 b cos (2 x)) · X + A COS (2 x) + b sin (2 x) y ~ "" \u003d ((- 2 a sin (2 x) + 2 b cos (2 x)) · x + a COS (2 x) + b sin (2 x)) "\u003d \u003d (- 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) \u003d \u003d (- 4 A cos (2 x) - 4 b sin (2 x)) · x - 4 a sin (2 x) + 4 b cos (2 x)

Then it can be seen that

y ~ "" + 4 y ~ \u003d 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 \u003d cos (2 x) + 3 sin (2 x) ⇔ - 4 A SIN (2 x) + 4 B cos (2 x) \u003d cos (2 x) + 3 sin (2 x)

It is necessary to equate the coefficients of sinuses and cosine. We get a system of type:

4 a \u003d 3 4 b \u003d 1 ⇔ a \u003d - 3 4 b \u003d 1 4

It follows that y ~ \u003d (a COS (2 x) + b sin (2 x) · x \u003d - 3 4 cos (2 x) + 1 4 sin (2 x) · x.

Answer:the general solution of the initial LFD of the second order with constant coefficients is considered

y \u003d Y 0 + Y ~ \u003d C 1 COS (2 x) + C 2 SIN (2 x) + - 3 4 COS (2 x) + 1 4 sin (2 x) · x

When f (x) \u003d eax · p n (x) sin (β x) + q k (x) cos (β x), then y ~ \u003d eax · (L m (x) sin (β x) + n m (x) cos (β x) · x γ. We have that R is the number of comprehensively conjugate pairs of roots relating 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, t, m, where M \u003d M a x (n, k). Finding coefficients L M (X) and N M (x) It is performed based on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).

Example 4.

Find a general solution y "" + 3 y "+ 2 y \u003d - E 3 x · ((38 x + 45) sin (5 x) + (8 x - 5) COS (5 x)).

Decision

Under the condition it can be seen that

α \u003d 3, β \u003d 5, p n (x) \u003d - 38 x - 45, q k (x) \u003d - 8 x + 5, n \u003d 1, k \u003d 1

Then m \u003d m a x (n, k) \u003d 1. We produce the finding Y 0, pre-writing the characteristic equation of the form:

k 2 - 3 k + 2 \u003d 0 d \u003d 3 2 - 4 · 1 · 2 \u003d 1 k 1 \u003d 3 - 1 2 \u003d 1, k 2 \u003d 3 + 1 2 \u003d 2

Received that the roots are valid and different. Hence y 0 \u003d C 1 E x + C 2 E 2 x. Next, it is necessary to look for a general decision, based on the inhomogeneous equation y ~ type

y ~ \u003d e α x · (L m (x) sin (β x) + n m (x) cos (β x) · x γ \u003d e 3 x · ((a x + b) COS (5 x) + (C x + d) sin (5 x)) · x 0 \u003d \u003d 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 \u003d 0, because there is no pair of conjugate roots belonging to the characteristic equation with α ± i β \u003d 3 ± 5 · i. These coefficients find from the equality obtained:

y ~ "" - 3 y ~ "+ 2 y ~ \u003d - 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))) \u003d - E 3 x ((38 x + 45) sin (5 x) + (8 x - 5) COS (5 x))

Finding derivative and similar components 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)) \u003d \u003d - E 3 x · (38 · x · sin (5 x) + 45 · SIN (5 x) + + 8 · x · COS (5 x) - 5 · COS (5 x))

After equating coefficients, we get a system of type

15 A + 23 C \u003d 38 10 A + 15 B - 3 C + 23 D \u003d 45 23 A - 15 C \u003d 8 - 3 A + 23 B - 10 C - 15 D \u003d - 5 ⇔ A \u003d 1 B \u003d 1 C \u003d 1 d \u003d 1

It follows from everything that

y ~ \u003d E 3 x · ((a x + b) cos (5 x) + (C x + d) sin (5 x)) \u003d \u003d E 3 x · ((x + 1) COS (5 x) + (x + 1) sin (5 x))

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

y \u003d y 0 + y ~ \u003d \u003d c 1 e x + c 2 e 2 x + e 3 x · ((x + 1) cos (5 x) + (x + 1) sin (5 x))

Algorithm Decision LDNU

Definition 1.

Any other type of function f (x) to solve compliance with the solution algorithm:

finding a general solution of the corresponding linear homogeneous equation, where Y 0 \u003d C 1 ⋅ Y 1 + C 2 ⋅ Y 2, where Y 1. and Y 2.are linearly independent private solutions With 1. and With 2are considered arbitrary constant;
the adoption as a general solution of the LFD y \u003d C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2;
determination of derivatives through a system of type C 1 "(x) + y 1 (x) + C 2" (x) · y 2 (x) \u003d 0 C 1 "(x) + y 1" (x) + C 2 " (x) · y 2 "(x) \u003d f (x), and finding functions C 1 (X) and C 2 (x) by integration.

Example 5.

Find a general solution for Y "" + 36 y \u003d 24 sin (6 x) - 12 COS (6 x) + 36 E 6 x.

Decision

We turn to the writing of the characteristic equation, pre-writing Y 0, Y "" + 36 y \u003d 0. We write and resolve:

k 2 + 36 \u003d 0 k 1 \u003d 6 i, k 2 \u003d - 6 i ⇒ y 0 \u003d C 1 cos (6 x) + c 2 sin (6 x) ⇒ y 1 (x) \u003d cos (6 x), Y 2 (x) \u003d sin (6 x)

We have that the recording of a general solution of a given equation will be viewed y \u003d C 1 (x) · COS (6 x) + C 2 (x) · sin (6 x). It is necessary to go to the definition of derived functions. C 1 (X) and C 2 (x) By system with equations:

C 1 "(x) · COS (6 x) + C 2" (x) · sin (6 x) \u003d 0 C 1 "(x) · (COS (6 x))" + C 2 "(x) · (SIN (6 x)) "\u003d 0 ⇔ C 1" (X) · COS (6 x) + C 2 "(x) · sin (6 x) \u003d 0 C 1" (x) (- 6 sin (6 x) + C 2 "(x) (6 cos (6 x)) \u003d 24 sin (6 x) - 12 cos (6 x) + 36 E 6 x

It is necessary to make a decision regarding C 1 "(x) and C 2 "(x) Using any way. Then 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 equations should be integrated. Then we write the resulting equations:

C 1 (x) \u003d 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) \u003d - 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 decision will look at:

y \u003d 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) \u003d \u003d - 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 \u003d y 0 + y ~ \u003d - 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)

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Here we apply the method of variation of permanent Lagrange to solve linear inhomogeneous second-order differential equations. Detailed description This method for solving the random order equations is set out on the page.
Solution of linear inhomogeneous differential equations of higher orders of Lagrange method \u003e\u003e\u003e.

Example 1.

Solve the second-order differential equation with constant coefficients by variation of permanent Lagrange:
(1)

Decision

Initially, we solve a homogeneous differential equation:
(2)

This is the second order equation.

We solve the square equation:
.
Roots multiples :. Fundamental system Solutions of equation (2) has the form:
(3) .
From here we get a general solution of a homogeneous equation (2):
(4) .

Various constant C. 1 and C. 2 . That is, we will replace in (4) permanent and functions:
.
We are looking for the solution of the initial equation (1) in the form:
(5) .

Find a derivative:
.
We connect functions and equation:
(6) .
Then
.

We find the second derivative:
.
We substitute in the initial equation (1):
(1) ;

.
Since and satisfy a homogeneous equation (2), the sum of members in each column of the last three lines gives zero and the previous equation acquires the form:
(7) .
Here .

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

Solving system of equations

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

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

.
By crawler formulas, we find:
;
.

So, we found derived functions:
;
.
We integrate (see root integration methods). Making substitution
; ; ; .

.
.

;
.

Answer

Example 2.

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

Decision

Step 1. Solution of a homogeneous equation

We solve a homogeneous differential equation:

(9)
We are looking for a decision in the form. We compile a characteristic equation:

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

Step 2. Variation of Permanent - Replacing Permanent Functions

Now vary constant C 1 and C. 2 . That is, replace in (11) permanent functions:
.
We are looking for the solution of the initial equation (8) as:
(12) .

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

Solving system of equations

We solve this system. We repel the expressions of 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 crawler formulas, we find:
;
.

.
Because, the module sign under the logarithm sign can be omitted. Multiply the numerator and denominator on:
.
Then
.

General solution of the original equation:

.

Linear homogeneous second-order differential equation with constant coefficients It has a general solution
where and linear independent private solutions of this equation.

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

Characteristic roots equations	Type of general solution
Roots and valid and different
Roots == valid and the same
Roings are complex ,

Example

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

Decision:
.

Having decided to find the roots
,
valid and different. Consequently, the general solution has the form:
.

Decision: Make a characteristic equation:
.

Having decided to find the roots

valid and identical. Consequently, the general solution has the form:
.

Decision: Make a characteristic equation:
.

Having decided to find the roots
complex. Consequently, the general solution has the form:.

Linear inhomogeneous second-order differential equation with constant coefficientshas appearance

Where
. (1)

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

View of a private solution
inhomogeneous equation (1) depending on the right side
:

Right part	Private solution
-Mogenous degree	where - The number of roots of the characteristic equation equal to zero.
	where = it is the root of the characteristic equation.
	Where - number, equal number the roots of the characteristic equation coinciding with .
	where - the number of roots of the characteristic equation coinciding with .

Consider various types of right parts of a linear inhomogeneous differential equation:

1.
where there is a numerous degree . Then a particular solution
can be sought in the form
where

, but - The number of roots of the characteristic equation equal to zero.

Example

Find a general solution
.

Decision:

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 a private solution we are looking for in the form where and - Unknown coefficients. Differentiating twice
and substituting
,
and
in the original equation, we find.

Equating coefficients with the same degrees in both parts of equality
,
Find
,
. So, the particular solution of this equation has the form
, and its overall solution.

2. Let the right-handed part
where there is a numerous degree . Then a particular solution
can be sought in the form
where
- a polynomial of the same extent as
, but - a number showing how many times it is the root of the characteristic equation.

Example

Find a general solution
.

Decision:

A) We will find a general solution of the corresponding homogeneous equation
. To do this, write the characteristic equation
. Find the roots of the last equation
. Consequently, the general solution of a homogeneous equation has the form
.

characteristic equation

where - Unknown coefficient. Differentiating twice
and substituting
,
and
in the original equation, we find. From
, i.e
or
.

So, the particular solution of this equation has the form
, and his general decision
.

3. Let the right-hand side relates to where
and - data numbers. Then a particular solution
can be sought in the form where and - Unknown odds, and - a number equal to the number of roots of the characteristic equation coinciding with
. If in the expression function
includes at least one of the functions
or
, then B.
we must always enter bothfunctions.

Example

Find a general solution.

Decision:

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

Where and - Unknown coefficients. Differentiating twice, get. Substituting
,
and
in the original equation, we find

Leading similar components, get

We equate the coefficients for
and
in the right and left parts of the equation, respectively. We get the system
. Solving it, find
,
.

So, the private solution of the initial differential equation has the form.

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

Establishment of Education "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Methodical instructions

according to the topics of the "linear differential equations of second order" by students of the accounting faculty of correspondence form of education (NEPO)

Gorki, 2013.

Linear differential equations

second order with constantcoefficients

Linear homogeneous differential equations

Linear second-order differential equation with constant coefficients Called the view equation

those. The equation that contains the desired function and its derivatives only in the first degree and does not contain their works. In this equation and
- Some numbers, and the function
set at some interval
.

If a
at the interval
, then equation (1) will take a view

, (2)

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

Consider a comprehensive function

, (3)

where
and
- Valid functions. If the function (3) is a comprehensive solution of equation (2), then the actual part
and imaginary part
solutions
separately are the solutions of the same homogeneous equation. Thus, any comprehensive solution of equation (2) generates two valid solutions to this equation.

Solutions of a homogeneous linear equation have properties:

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

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

If a and there are solutions of equation (2), then their linear combination
will also be a solution equation (2), where and
- Arbitrary constant.

Functions
and
called linearly dependent At the interval
if there are such numbers and
, not equal zero At the same time, that equality is performed on this interval.

If equality (4) takes place only when
and
, then functions
and
called linearly independent At the interval
.

Example 1. . Functions
and
linearly dependent because
on the whole numeric straight. In this example
.

Example 2. . Functions
and
linearly independent at any interval, because equality
possible only in the case when
, I.
.

Building a general solution of linear homogeneous

equations

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

Linearly independent solutions of equation (2) will be signed as

, (5)

where - Some number. Then
,
. Substitute these expressions in equation (2):

or
.

As
T.
. Thus, the function
will be the solution of equation (2) if will satisfy the equation

. (6)

Equation (6) is called characteristic equation For equation (2). This equation is an algebraic square equation.

Let be and there are roots of this equation. They can be either valid and different, or complex, or valid and equal. Consider these cases.

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

where and
- Arbitrary constant.

Example 3.
.

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

Integrated number called the expression of the view
where and - Actual numbers, and
called an imaginary unit. If a
, then
called purely imaginary. If
, then
identified with a valid number .

Number called a valid part of the integrated number, and - imaginary part. If two complex numbers differ from each other only by the image of the imaginary part, they are closed conjugate:
,
.

Example 4. . Solve square equation
.

Decision . Discriminant equation
. Then. Similarly,
. Thus, this square equation has conjugate complex roots.

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

,
.

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

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

where and
- Arbitrary constant.

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

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

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

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

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

Inhomogeneous linear second-order differential equations with constant coefficients

and a special right

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

In some cases, the private solution of the inhomogeneous equation can be found quite just in appearance of the right
equations (1). Consider cases when possible.

those. The right side of the inhomogeneous equation is a polynomial degree m.. If a
it is not the root of the characteristic equation, the particular solution of the inhomogeneous equation should be seen as a polynomial m..

Factors
defined in the process of finding a private solution.

If
it is the root of the characteristic equation, the private solution of the inhomogeneous equation should be sought in the form of

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

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

As
it is not a root of a characteristic equation, the particular solution of the inhomogeneous equation will be signed as a function
. Find derivatives of this function
,
and we substitute them in this equation:

or . We equate the coefficients for and free members:
Solving this system, get
,
. Then the particular solution of the inhomogeneous equation has the form
And the general solution of this inhomogeneous equation will be the sum of the overall solution of the corresponding homogeneous equation and a particular solution of heterogeneous:
.

Let the heterogeneous equation relates

If a
it is not a root of the characteristic equation, the particular solution of the inhomogeneous equation should be signed in the form. If
there is a root of the characteristic equation of multiplicity k. (k.\u003d 1 or k.\u003d 2), in this case, the particular solution of the inhomogeneous equation will be viewed.

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

Decision . The characteristic equation for the corresponding homogeneous equation has the form
. His roots
,
. In this case, the general solution of the corresponding homogeneous equation is recorded as
.

Since the number 3 is not the root of the characteristic equation, the particular solution of the inhomogeneous equation should be sought in the form of
. Find the derivatives of the first and second orders :,

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

We equate the coefficients for and free members:

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

Lagrange Method Variations Arbitrary Permanent

The method of variation of arbitrary constants can be applied to any non-uniform linear equation with constant coefficients, regardless of the type of the right part. This method allows you to always find a general solution of an inhomogeneous equation, if a general solution is known to the corresponding homogeneous equation.

Let be
and
are linearly independent solutions of equation (2). Then the general solution of this equation is
where and
- Arbitrary constant. The essence of the method of variation of arbitrary constants is that the general solution of the equation (1) is searched as

where
and
- New unknown features that need to be found. Since unknown functions are two, then there are two equations containing these functions for their findings. These two equations make up the system

which is a linear algebraic system of equations relative to
and
. Solving this system, we will find
and
. Integrating both parts of the equalities received, find

and
.

Substituting these expressions in (9), we obtain a general solution of a non-uniform linear equation (1).

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

Decision. The characteristic equation for a homogeneous equation corresponding to this differential equation is
. The roots of its complex
,
. As
and
T.
,
, and the general solution of a 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 is

Deciding this system, we will find
,
. Then

,
. Substitute the received expressions in the general decision formula:

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

Questions for self-controlling knowledge

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

What a linear differential equation is called homogeneous, and what is heterogeneous?

What properties is the 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 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 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 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 private solution of a linear inhomogeneous equation, if the roots of the characteristic equation are different and are not equal to zero, and the right side of the equation is a polynomial degree m.?

In what form is a private solution of a linear inhomogeneous equation, if among the roots of the characteristic equation there is one zero, and the right part of the equation is a polynomial m.?

What is the essence of the Lagrange method?