In some problems of physics, it is not possible to establish a direct connection between the quantities describing the process. But it is possible to obtain an equality containing the derivatives of the functions under study. This is how differential equations arise and the need to solve them to find an unknown function.

This article is intended for those who are faced with the problem of solving a differential equation in which the unknown function is a function of one variable. The theory is structured in such a way that with zero knowledge of differential equations, you can cope with your task.

Each type differential equations the solution method is put in accordance with detailed explanations and solutions to typical examples and problems. All you have to do is determine the type of differential equation of your problem, find a similar analyzed example and carry out similar actions.

For successful solution differential equations, you will also need the ability to find sets of antiderivatives ( indefinite integrals) various functions. If necessary, we recommend that you refer to the section.

First, we will consider the types of ordinary differential equations of the first order that can be resolved with respect to the derivative, then we will move on to second-order ODEs, then we will dwell on higher-order equations and end with systems of differential equations.

Recall that if y is a function of the argument x.

First order differential equations.

The simplest differential equations of the first order of the form.

Let's write down a few examples of such remote control .

Differential equations can be resolved with respect to the derivative by dividing both sides of the equality by f(x) . In this case, we arrive at an equation that will be equivalent to the original one for f(x) ≠ 0. Examples of such ODEs are .

If there are values ​​of the argument x at which the functions f(x) and g(x) simultaneously vanish, then additional solutions appear. Additional solutions to the equation given x are any functions defined for these argument values. Examples of such differential equations include:

Second order differential equations.

Linear homogeneous differential equations of the second order with constant coefficients.

LDE with constant coefficients is a very common type of differential equation. Their solution is not particularly difficult. First the roots are found characteristic equation . For different p and q, three cases are possible: the roots of the characteristic equation can be real and different, real and coinciding or complex conjugates. Depending on the values ​​of the roots of the characteristic equation, the general solution of the differential equation is written as , or , or respectively.

For example, consider a linear homogeneous second-order differential equation with constant coefficients. The roots of its characteristic equation are k 1 = -3 and k 2 = 0. The roots are real and different, therefore, the general solution of the LODE with constant coefficients has the form


Linear inhomogeneous differential equations of the second order with constant coefficients.

The general solution of a second-order LDDE with constant coefficients y is sought in the form of the sum general solution corresponding to LOD and a particular solution to the original one is not homogeneous equation, that is, . The previous paragraph is devoted to finding a general solution to a homogeneous differential equation with constant coefficients. And a particular solution is determined either by the method of indefinite coefficients with a certain form function f(x) on the right side of the original equation, or by the method of varying arbitrary constants.

As examples of second-order LDDEs with constant coefficients, we give


Understand the theory and become familiar with detailed solutions We offer you examples on the page of linear inhomogeneous differential equations of the second order with constant coefficients.

Linear homogeneous differential equations (LODE) and linear inhomogeneous differential equations (LNDEs) of the second order.

A special case of differential equations of this type are LODE and LDDE with constant coefficients.

The general solution of the LODE on a certain segment is represented by a linear combination of two linearly independent partial solutions y 1 and y 2 of this equation, that is, .

The main difficulty lies precisely in finding linearly independent partial solutions to a differential equation of this type. Typically, particular solutions are selected from the following systems linearly independent functions:


However, particular solutions are not always presented in this form.

An example of a LOD is .

The general solution of the LDDE is sought in the form , where is the general solution of the corresponding LDDE, and is the particular solution of the original differential equation. We just talked about finding it, but it can be determined using the method of varying arbitrary constants.

An example of LNDU can be given .

Differential equations of higher orders.

Differential equations that allow a reduction in order.

Order of differential equation , which does not contain the desired function and its derivatives up to k-1 order, can be reduced to n-k by replacing .

In this case, the original differential equation will be reduced to . After finding its solution p(x), it remains to return to the replacement and determine the unknown function y.

For example, the differential equation after the replacement, it will become an equation with separable variables, and its order will be reduced from third to first.

This article addresses the issue of solving linear inhomogeneous second-order differential equations with constant coefficients. The theory will be discussed along with examples of given problems. To decipher unclear terms, it is necessary to refer to the topic about the basic definitions and concepts of the theory of differential equations.

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

Let us move on to the formulation of the theorem for the general solution of the LNDE.

Yandex.RTB R-A-339285-1

General solution theorem for LDNU

Theorem 1

A 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 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 general solution y 0, which corresponds to the LOD and some particular solution y ~, where the original inhomogeneous equation is y = y 0 + y ~.

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

The choice of a particular solution to the LPDE 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 second-order differential equations 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 LPDE is found using 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 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 Cauchy's theorem y "" - 2 y " = x 2 + 1 , y (0) = 2 , y " (0) = 1 4 .

Solution

In other words, it is necessary to move on 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 the linear inhomogeneous equation is the sum of the general solution that corresponds to the equation y 0 or a particular solution to the inhomogeneous equation y ~, that is, y = y 0 + y ~.

First, we will find a general solution for the LNDU, and then a particular one.

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

k 2 - 2 k = 0 k (k - 2) = 0 k 1 = 0 , k 2 = 2

We found that the roots are different and real. Therefore, let's write down

y 0 = C 1 e 0 x + C 2 e 2 x = C 1 + C 2 e 2 x.

Let's find y ~ . It can be seen that the right side behind given equation is a polynomial of the second degree, then one of the roots is equal to zero. From this we obtain that a particular solution for y ~ will be

y ~ = Q 2 (x) x γ = (A x 2 + B x + C) x = A x 3 + B x 2 + C x, where the values ​​of A, B, C take on undetermined 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 of x, we obtain a system of linear expressions - 6 A = 1 6 A - 4 B = 0 2 B - 2 C = 1. When solving by any of the methods, we will find the coefficients and write: A = - 1 6, B = - 1 4, C = - 3 4 and y ~ = A x 3 + B x 2 + C x = - 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 necessary to determine the values C 1 And C 2, based on an equality of the form y = 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 Cauchy's 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 the product of a polynomial with degree n and an exponent f (x) = P n (x) · e a x , then we obtain that a particular solution of the second-order LPDE 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 to a differential equation of the form y "" - 2 y " = (x 2 + 1) · e x .

Solution

The equation general view y = y 0 + y ~ . The above equation corresponds to LOD y "" - 2 y " = 0. From the previous example it can be seen that its roots are equal k 1 = 0 and k 2 = 2 and y 0 = C 1 + C 2 e 2 x by the characteristic equation.

It's clear that right side equation is x 2 + 1 · e x . From here the LPDE is found through y ~ = e a x · Q n (x) · x γ, where Q n (x) 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. From here 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 that 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 with 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 is clear 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 the LNDDE, and y = y 0 + y = C 1 e 2 x - e x · x 2 + 3 - a general solution for a second-order inhomogeneous dif equation.

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 a partial solution of the LPDE is considered to be an equation of the form y ~ = A cos β x + B sin β x · x γ, where A and B are considered undetermined coefficients, and r is 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 using the equality y ~ "" + p · y ~ " + q · y ~ = f (x).

Example 3

Find the general solution to 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 = 0 k 2 = - 4 k 1 = 2 i , k 2 = - 2 i

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

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

The roots of the characteristic equation are considered to be the 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 We will look for coefficients A and B 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 clear 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) + 4 B cos (2 x) = cos (2 x) + 3 sin (2 x)

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 second-order LDDE with constant coefficients is considered

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 Nm(x) are polynomials of degree n, k, m, m, where m = m a x (n, k). Finding coefficients Lm(x) And Nm(x) is made 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

According to the condition it is clear 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 a 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 the 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. We find these coefficients 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 everything 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 (5 x))

Answer: Now we have obtained a general solution to 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 type of function f (x) for solution requires compliance with the solution algorithm:

• finding a general solution to the corresponding linear homogeneous equation, where y 0 = C 1 ⋅ y 1 + C 2 ⋅ y 2, where y 1 And y 2 are linearly independent partial solutions of the LODE, C 1 And C 2 are considered arbitrary constants;
• accepting as a general solution the LNDE y = C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2 ;
• determination 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 general solution of the given equation will be written as y = C 1 (x) · cos (6 x) + C 2 (x) · sin (6 x) . It is necessary to move on to the definition of derivative functions C 1 (x) And C2(x) according to a 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)) = = 24 sin (6 x) - 12 cos (6 x) + 36 e 6 x

A decision needs to be made regarding C 1" (x) And C 2" (x) using any method. Then we write:

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

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Differential equations of second order and higher orders.
Linear differential equations of the second order with constant coefficients.
Examples of solutions.

Let's move on to considering second-order differential equations and higher-order differential equations. If you have a vague idea of ​​what a differential equation is (or don’t understand what it is at all), then I recommend starting with the lesson First order differential equations. Examples of solutions. Many solution principles and basic concepts first order diffusers automatically extend to differential equations of higher orders, therefore it is very important to first understand the first order equations.

Many readers may have a prejudice that remote control of the 2nd, 3rd and other orders is something very difficult and inaccessible to master. This is wrong . Learn to solve diffuses higher order hardly more complicated than “ordinary” 1st order DEs. And in some places it is even simpler, since the solutions actively use material from the school curriculum.

Most Popular second order differential equations. To a second order differential equation Necessarily includes the second derivative and not included

It should be noted that some of the babies (and even all of them at once) may be missing from the equation; it is important that the father is at home. The most primitive second-order differential equation looks like this:

Third order differential equations in practical tasks are much less common, according to my subjective observations in State Duma they would get about 3-4% of the vote.

To a third order differential equation Necessarily includes the third derivative and not included derivatives of higher orders:

The simplest third-order differential equation looks like this: – dad is at home, all the children are out for a walk.

In a similar way, you can define differential equations of the 4th, 5th and higher orders. In practical problems, such control systems rarely fail, however, I will try to give relevant examples.

Higher order differential equations, which are proposed in practical problems, can be divided into two main groups.

1) The first group - the so-called equations that can be reduced in order. Come on!

2) Second group – linear equations higher orders with constant coefficients. Which we will start looking at right now.

Linear differential equations of the second order
with constant coefficients

In theory and practice, two types of such equations are distinguished: homogeneous equation And inhomogeneous equation.

Homogeneous second order DE with constant coefficients It has next view:
, where and are constants (numbers), and on the right side – strictly zero.

As you can see, there are no particular difficulties with homogeneous equations, the main thing is decide correctly quadratic equation .

Sometimes there are non-standard homogeneous equations, for example an equation in the form , where at the second derivative there is some constant different from unity (and, naturally, different from zero). The solution algorithm does not change at all; you should calmly compose a characteristic equation and find its roots. If the characteristic equation will have two different real roots, for example: , then the general solution will be written according to the usual scheme: .

In some cases, due to a typo in the condition, “bad” roots may result, something like . What to do, the answer will have to be written like this:

With “bad” conjugate complex roots like no problem either, general solution:

That is, there is a general solution anyway. Because any quadratic equation has two roots.

In the final paragraph, as I promised, we will briefly consider:

Linear homogeneous equations of higher orders

Everything is very, very similar.

A linear homogeneous equation of third order has the following form:
, where are constants.
For this equation, you also need to create a characteristic equation and find its roots. The characteristic equation, as many have guessed, looks like this:
, and it Anyway It has exactly three root

Let, for example, all roots be real and distinct: , then the general solution will be written as follows:

If one root is real, and the other two are conjugate complex, then we write the general solution as follows:

A special case, when all three roots are multiples (the same). Let's consider the simplest homogeneous DE of the 3rd order with a lonely father: . The characteristic equation has three coincident zero roots. We write the general solution as follows:

If the characteristic equation has, for example, three multiple roots, then the general solution, accordingly, is as follows:

Example 9

Solve a homogeneous third order differential equation

Solution: Let's compose and solve the characteristic equation:

, – one real root and two conjugate complex roots are obtained.

Answer: common decision

Similarly, we can consider a fourth-order linear homogeneous equation with constant coefficients: , where are constants.

Linear differential equation of the second order called an equation of the form

y"" + p(x)y" + q(x)y = f(x) ,

Where y is the function to be found, and p(x) , q(x) And f(x) - continuous functions on a certain interval ( a, b) .

If the right side of the equation is zero ( f(x) = 0), then the equation is called linear homogeneous equation . The practical part of this lesson will mainly be devoted to such equations. If the right side of the equation is not equal to zero ( f(x) ≠ 0), then the equation is called .

In the problems we are required to solve the equation for y"" :

y"" = −p(x)y" − q(x)y + f(x) .

Second order linear differential equations have a unique solution Cauchy problems .

Linear homogeneous differential equation of the second order and its solution

Consider a linear homogeneous second order differential equation:

y"" + p(x)y" + q(x)y = 0 .

If y1 (x) And y2 (x) are particular solutions of this equation, then the following statements are true:

1) y1 (x) + y 2 (x) - is also a solution to this equation;

2) Cy1 (x) , Where C- an arbitrary constant (constant), is also a solution to this equation.

From these two statements it follows that the function

C1 y 1 (x) + C 2 y 2 (x)

is also a solution to this equation.

A fair question arises: is this solution general solution of a linear homogeneous differential equation of the second order , that is, such a solution in which, for different values C1 And C2 Is it possible to get all possible solutions to the equation?

The answer to this question is: maybe, but under certain conditions. This condition on what properties particular solutions should have y1 (x) And y2 (x) .

And this condition is called condition linear independence private solutions.

Theorem. Function C1 y 1 (x) + C 2 y 2 (x) is a general solution to a linear homogeneous second order differential equation if the functions y1 (x) And y2 (x) linearly independent.

Definition. Functions y1 (x) And y2 (x) are called linearly independent if their ratio is a constant non-zero:

y1 (x)/y 2 (x) = k ; k = const ; k ≠ 0 .

However, determining by definition whether these functions are linearly independent is often very laborious. There is a way to establish linear independence using the Wronski determinant W(x) :

If the Wronski determinant is not equal to zero, then the solutions are linearly independent . If the Wronski determinant is zero, then the solutions are linearly dependent.

Example 1. Find the general solution of a linear homogeneous differential equation.

Solution. We integrate twice and, as is easy to see, in order for the difference between the second derivative of a function and the function itself to be equal to zero, the solutions must be associated with an exponential whose derivative is equal to itself. That is, the partial solutions are and .

Since the Wronski determinant

is not equal to zero, then these solutions are linearly independent. Therefore, the general solution to this equation can be written as

.

Linear homogeneous second order differential equations with constant coefficients: theory and practice

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

y"" + py" + qy = 0 ,

Where p And q- constant values.

The fact that this is a second-order equation is indicated by the presence of the second derivative of the desired function, and its homogeneity is indicated by a zero on the right side. The values ​​already mentioned above are called constant coefficients.

To solve a linear homogeneous second order differential equation with constant coefficients , you must first solve the so-called characteristic equation of the form

k² + pq + q = 0 ,

which, as can be seen, is an ordinary quadratic equation.

Depending on the solution of the characteristic equation, three different options are possible solutions to a linear homogeneous second order differential equation with constant coefficients , which we will now analyze. For complete definiteness, we will assume that all particular solutions have been tested by the Wronski determinant and it is not equal to zero in all cases. Those who doubt it, however, can check this themselves.

The roots of the characteristic equation are real and distinct

In other words, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form

.

Example 2. Solve a linear homogeneous differential equation

.

Example 3. Solve a linear homogeneous differential equation

.

Solution. The characteristic equation has the form , its roots and are real and distinct. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

.

The roots of the characteristic equation are real and equal

That is, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form

.

Example 4. Solve a linear homogeneous differential equation

.

Solution. Characteristic equation has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

Example 5. Solve a linear homogeneous differential equation

.

Solution. The characteristic equation has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form

Consider a linear homogeneous differential equation with constant coefficients:
(1) .
Its solution can be obtained by following general method reduction of order.

However, it is easier to immediately obtain the fundamental system n linearly independent solutions and based on it create a general solution. In this case, the entire solution procedure is reduced to the following steps.

We are looking for a solution to equation (1) in the form . We get:
(2) .
characteristic equation
(3) .
It has n roots. We solve equation (2) and find its roots.
(4) .

Then the characteristic equation (2) can be represented in the following form:

Each root corresponds to one of the linearly independent solutions of the fundamental system of solutions to equation (1). Then the general solution to the original equation (1) has the form: Real roots
.

Let's consider real roots
. Let the root be single. That is, the factor enters the characteristic equation (3) only once. Then this root corresponds to the solution
.
Let be a multiple root of multiplicity p.
; ; ; ...; .

That is

. In this case, the multiplier is p times:
.
These multiple (equal) roots correspond to p linearly independent solutions of the original equation (1):
.

Complex roots
; .

Consider complex roots
.
. Let us express the complex root in terms of the real and imaginary parts: Since the coefficients of the original are real, then in addition to the root there is a complex conjugate root Let the complex root be multiple. Then a pair of roots corresponds to two linearly independent solutions: Since the coefficients of the original are real, then in addition to the root there is a complex conjugate root linearly independent solutions:
; ; ; ... ;
; ; ; ... .

After fundamental system Linearly independent solutions are found, and we obtain the general solution.

Examples of problem solutions

Example 1

Solve the equation:
.

Solution

.
Let's transform it:
;
;
.

Let's look at the roots of this equation. We got four complex roots of multiplicity 2:
; .
They correspond to four linearly independent solutions of the original equation:
; ; ; .

We also have three real roots of multiple 3:
.
They correspond to three linearly independent solutions:
; ; .

The general solution to the original equation has the form:
.

Answer

Example 2

Solve the equation

Solution

We are looking for a solution in the form .
.
We compose the characteristic equation:
.

Solving a quadratic equation.
.
We got two complex roots:
.
They correspond to two linearly independent solutions:
.