Linear homogeneous equation of the second order. Second order and higher order differential equations

This article reveals the question of solving linear inhomogeneous differential equations of the second order with constant coefficients. The theory will be considered along with examples of the given problems. To decipher 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 a 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 can be seen that the right side of the given equation is a polynomial of the second degree, 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

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

It can be seen that the 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 has been obtained:

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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Fundamentals of solving linear inhomogeneous differential equations of the second order (LNDE-2) with constant coefficients (PC)

A second-order CLDE with constant coefficients $p$ and $q$ has the form $y""+p\cdot y"+q\cdot y=f\left(x\right)$, where $f\left(x \right)$ is a continuous function.

The following two statements are true with respect to the 2nd LNDE with PC.

Assume that some function $U$ is an arbitrary particular solution of an inhomogeneous differential equation. Let us also assume that some function $Y$ is a general solution (OR) of the corresponding linear homogeneous differential equation (LODE) $y""+p\cdot y"+q\cdot y=0$. Then the OR of LNDE-2 is equal to the sum of the indicated private and general solutions, i.e. $y=U+Y$.

If the right side of the 2nd order LIDE is the sum of functions, that is, $f\left(x\right)=f_(1) \left(x\right)+f_(2) \left(x\right)+. ..+f_(r) \left(x\right)$, then first you can find the PD $U_(1) ,U_(2) ,...,U_(r) $ that correspond to each of the functions $f_( 1) \left(x\right),f_(2) \left(x\right),...,f_(r) \left(x\right)$, and after that write the LNDE-2 PD as $U=U_(1) +U_(2) +...+U_(r) $.

Solution of 2nd order LNDE with PC

Obviously, the form of one or another PD $U$ of a given LNDE-2 depends on the specific form of its right-hand side $f\left(x\right)$. The simplest cases of searching for the PD of LNDE-2 are formulated as the following four rules.

Rule number 1.

The right side of LNDE-2 has the form $f\left(x\right)=P_(n) \left(x\right)$, where $P_(n) \left(x\right)=a_(0) \cdot x^(n) +a_(1) \cdot x^(n-1) +...+a_(n-1) \cdot x+a_(n) $, that is, it is called a polynomial of degree $n$. Then its PR $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) $, where $Q_(n) \left(x\right)$ is another polynomial of the the same degree as $P_(n) \left(x\right)$, and $r$ is the number of zero roots of the characteristic equation of the corresponding LODE-2. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the method of indefinite coefficients (NC).

Rule number 2.

The right side of LNDE-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot P_(n) \left(x\right)$, where $P_(n) \left( x\right)$ is a polynomial of degree $n$. Then its PD $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) \cdot e^(\alpha \cdot x) $, where $Q_(n) \ left(x\right)$ is another polynomial of the same degree as $P_(n) \left(x\right)$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $\alpha $. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the NK method.

Rule number 3.

The right part of LNDE-2 has the form $f\left(x\right)=a\cdot \cos \left(\beta \cdot x\right)+b\cdot \sin \left(\beta \cdot x\right) $, where $a$, $b$ and $\beta $ are known numbers. Then its PD $U$ is searched for in the form $U=\left(A\cdot \cos \left(\beta \cdot x\right)+B\cdot \sin \left(\beta \cdot x\right)\right )\cdot x^(r) $, where $A$ and $B$ are unknown coefficients, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $i\cdot \beta $. The coefficients $A$ and $B$ are found by the NDT method.

Rule number 4.

The right side of LNDE-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot \left$, where $P_(n) \left(x\right)$ is a polynomial of degree $ n$, and $P_(m) \left(x\right)$ is a polynomial of degree $m$. Then its PD $U$ is searched for in the form $U=e^(\alpha \cdot x) \cdot \left\cdot x^(r) $, where $Q_(s) \left(x\right)$ and $ R_(s) \left(x\right)$ are polynomials of degree $s$, the number $s$ is the maximum of two numbers $n$ and $m$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2, equal to $\alpha +i\cdot \beta $. The coefficients of the polynomials $Q_(s) \left(x\right)$ and $R_(s) \left(x\right)$ are found by the NK method.

The NK method consists in applying the following rule. In order to find the unknown coefficients of the polynomial, which are part of the particular solution of the inhomogeneous differential equation LNDE-2, it is necessary:

• substitute the PD $U$, written in general form, into the left part of LNDE-2;
• on the left side of LNDE-2, perform simplifications and group terms with the same powers $x$;
• in the resulting identity, equate the coefficients of the terms with the same powers $x$ of the left and right sides;
• solve the resulting system of linear equations for unknown coefficients.

Example 1

Task: find the OR LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. Also find the PR , satisfying the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$.

Write the corresponding LODA-2: $y""-3\cdot y"-18\cdot y=0$.

Characteristic equation: $k^(2) -3\cdot k-18=0$. The roots of the characteristic equation: $k_(1) =-3$, $k_(2) =6$. These roots are real and distinct. Thus, the OR of the corresponding LODE-2 has the form: $Y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) $.

The right part of this LNDE-2 has the form $\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. It is necessary to consider the coefficient of the exponent of the exponent $\alpha =3$. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PR of this LNDE-2 has the form $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $.

We will look for the coefficients $A$, $B$ using the NK method.

We find the first derivative of the CR:

$U"=\left(A\cdot x+B\right)^((") ) \cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot \left( e^(3\cdot x) \right)^((") ) =$

$=A\cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot 3\cdot e^(3\cdot x) =\left(A+3\cdot A\ cdot x+3\cdot B\right)\cdot e^(3\cdot x) .$

We find the second derivative of the CR:

$U""=\left(A+3\cdot A\cdot x+3\cdot B\right)^((") ) \cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot \left(e^(3\cdot x) \right)^((") ) =$

$=3\cdot A\cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot 3\cdot e^(3\cdot x) =\left(6\cdot A+9\cdot A\cdot x+9\cdot B\right)\cdot e^(3\cdot x) .$

We substitute the functions $U""$, $U"$ and $U$ instead of $y""$, $y"$ and $y$ into the given LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x).$ At the same time, since the exponent $e^(3\cdot x) $ is included as a factor in all components, then its can be omitted.

$6\cdot A+9\cdot A\cdot x+9\cdot B-3\cdot \left(A+3\cdot A\cdot x+3\cdot B\right)-18\cdot \left(A\ cdot x+B\right)=36\cdot x+12.$

We perform actions on the left side of the resulting equality:

$-18\cdot A\cdot x+3\cdot A-18\cdot B=36\cdot x+12.$

We use the NC method. We get a system of linear equations with two unknowns:

$-18\cdot A=36;$

$3\cdot A-18\cdot B=12.$

The solution to this system is: $A=-2$, $B=-1$.

The CR $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $ for our problem looks like this: $U=\left(-2\cdot x-1\right) \cdot e^(3\cdot x) $.

The OR $y=Y+U$ for our problem looks like this: $y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) +\ left(-2\cdot x-1\right)\cdot e^(3\cdot x) $.

In order to search for a PD that satisfies the given initial conditions, we find the derivative $y"$ OR:

$y"=-3\cdot C_(1) \cdot e^(-3\cdot x) +6\cdot C_(2) \cdot e^(6\cdot x) -2\cdot e^(3\ cdot x) +\left(-2\cdot x-1\right)\cdot 3\cdot e^(3\cdot x) .$

We substitute in $y$ and $y"$ the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$:

$6=C_(1) +C_(2) -1; $

$1=-3\cdot C_(1) +6\cdot C_(2) -2-3=-3\cdot C_(1) +6\cdot C_(2) -5.$

We got a system of equations:

$C_(1) +C_(2) =7;$

$-3\cdot C_(1) +6\cdot C_(2) =6.$

We solve it. We find $C_(1) $ using Cramer's formula, and $C_(2) $ is determined from the first equation:

$C_(1) =\frac(\left|\begin(array)(cc) (7) & (1) \\ (6) & (6) \end(array)\right|)(\left|\ begin(array)(cc) (1) & (1) \\ (-3) & (6) \end(array)\right|) =\frac(7\cdot 6-6\cdot 1)(1\ cdot 6-\left(-3\right)\cdot 1) =\frac(36)(9) =4; C_(2) =7-C_(1) =7-4=3.$

Thus, the PD of this differential equation is: $y=4\cdot e^(-3\cdot x) +3\cdot e^(6\cdot x) +\left(-2\cdot x-1\right )\cdot e^(3\cdot x) $.

Differential equations of the second order and higher orders.
Linear DE of the second order with constant coefficients.
Solution examples.

We pass to the consideration of differential equations of the second order and differential equations of higher orders. 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. Solution examples. Many solution principles and basic concepts of first-order diffurs are automatically extended to higher-order differential equations, therefore it is very important to first understand the first order equations.

Many readers may have a prejudice that DE of the 2nd, 3rd, and other orders is something very difficult and inaccessible for mastering. This is not true . Learning to solve higher-order diffuses is hardly more difficult than “ordinary” 1st-order DEs. And in some places it is even easier, since the material of the school curriculum is actively used in the decisions.

Most Popular second order differential equations. Into 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 at once) may be missing from the equation, it is important that the father was 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 the State Duma, they would gain about 3-4% of the votes.

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

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

Similarly, differential equations of the 4th, 5th and higher orders can be defined. In practical problems, such DE slips extremely rarely, however, I will try to give relevant examples.

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

1) The first group - the so-called lower-order equations. Fly in!

2) The second group - higher-order linear equations with constant coefficients. Which we will begin to consider right now.

Second Order Linear Differential Equations
with constant coefficients

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

Homogeneous DE of the second order with constant coefficients has the following form:
, where and are constants (numbers), and on the right side - strictly zero.

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

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

In some cases, due to a typo in the condition, “bad” roots can turn out, 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, a general solution exists in any case. Because any quadratic equation has two roots.

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

Higher Order Linear Homogeneous Equations

Everything is very, very similar.

The linear homogeneous equation of the third order has the following form:
, where are constants.
For this equation, you also need to compose 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 can 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 is 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, respectively, is:

Example 9

Solve a homogeneous differential equation of the third order

Solution: We compose and solve the characteristic equation:

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

Answer: common decision

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