Find the general solution of the equation with constant coefficients. Linear differential equations of the second order

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Methodical instructions

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

Gorki, 2013

Linear differential equations

second order with constantscoefficients

Linear homogeneous differential equations

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

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

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

, (2)

and called linear homogeneous ... Otherwise, equation (1) is called linear non-uniform .

Consider the complex function

, (3)

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

Solutions to a homogeneous linear equation have the following properties:

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

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

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

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

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

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

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

Construction of a general solution to a linear homogeneous

equations

To find common decision equation (2), you need to find two of its linearly independent solutions and ... Linear combination of these solutions
, where and
- arbitrary constants, and will give a general solution to a linear homogeneous equation.

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

, (5)

where - some number. Then
,
... Substitute these expressions into equation (2):

or
.

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

. (6)

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

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

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

,

where and
- arbitrary constants.

Example 3
.

Solution ... The characteristic equation for this differential will be
... Having solved this quadratic equation, we find its roots
and
... Functions
and
are solutions to a differential equation. The general solution to this equation has the form
.

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

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

Example 4 ... Solve Quadratic Equation
.

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

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

,
.

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

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

where and
- arbitrary constants.

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

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

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

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

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

Inhomogeneous linear differential equations of the second order with constant coefficients

and special right side

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

In some cases, a particular solution to an inhomogeneous equation can be found quite simply by the form of the right-hand side
equation (1). Consider the cases where this is possible.

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

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

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

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

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

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

or . Let us equate the coefficients at and free members:
Having solved this system, we 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 general solution of the corresponding homogeneous equation and the particular solution of the inhomogeneous one:
.

Let the inhomogeneous equation have the form

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

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

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

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

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

Let us equate the coefficients at and free members:

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

.

Lagrange's method of variation of arbitrary constants

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

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

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

which is a linear algebraic system of equations for
and
... Solving this system, we find
and
... Integrating both sides of the obtained equalities, we find

and
.

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

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

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

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

Having solved this system, we will find
,
... Then

,
... Substitute the obtained expressions into the general solution formula:

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

Questions for self-control of knowledge

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

Which linear differential equation is called homogeneous and which is called inhomogeneous?

What properties does a linear homogeneous equation have?

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

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

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

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

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

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

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

What is the essence of the Lagrange method?

We have made sure that, in the case when the general solution of a linear homogeneous equation is known, it is possible to find a general solution of the inhomogeneous equation by the method of variation of arbitrary constants. However, the question of how to find a general solution to the homogeneous equation remained open. In the special case when in the linear differential equation (3) all the coefficients p i(X)= a i - constants, it can be solved quite simply, even without integration.

Consider a linear homogeneous differential equation with constant coefficients, i.e., equations of the form

y (n) + a 1 y (n – 1) + ... a n – 1 y " + a n y = 0, (14)

where and i- constants (i= 1, 2, ...,n).

As is known, for a linear homogeneous equation of the 1st order, the solution is a function of the form e kx. We will seek a solution to equation (14) in the form j (X) = e kx.

Let us substitute into equation (14) the function j (X) and its derivatives of order m (1 £ m£ n)j (m) (X) = k m е kx... We get

(k n + a 1 k n – 1 + ... a n – 1 k + a n)e kx = 0,

but e k x ¹ 0 for any X, That's why

k n + a 1 k n – 1 + ... a n – 1 k + a n = 0. (15)

Equation (15) is called characteristic equation, the polynomial on the left,- characteristic polynomial , its roots- characteristic roots differential equation (14).

Conclusion:

functionj (X) = e kx - the solution of the linear homogeneous equation (14) if and only if the number k - the root of the characteristic equation (15).

Thus, the process of solving the linear homogeneous equation (14) is reduced to solving the algebraic equation (15).

Various cases of characteristic roots are possible.

1.All roots of the characteristic equation are real and different.

In this case n different characteristic roots k 1 ,k 2 ,..., k n corresponds to n different solutions of homogeneous equation (14)

It can be shown that these solutions are linearly independent, and therefore form fundamental system solutions. Thus, the general solution to the equation is the function

where WITH 1 , C 2 , ..., С n - arbitrary constants.

EXAMPLE 7. Find the general solution to a linear homogeneous equation:

a) at¢ ¢ (X) - 6at¢ (X) + 8at(X) = 0, b) at¢ ¢ ¢ (X) + 2at¢ ¢ (X) - 3at¢ (X) = 0.

Solution. Let's compose a characteristic equation. To do this, we replace the derivative of the order m functions y(x) to the appropriate degree

k(at (m) (x) « k m),

the function itself at(X) as the zero-order derivative is replaced by k 0 = 1.

In case (a), the characteristic equation has the form k 2 - 6k + 8 = 0. The roots of this quadratic equation k 1 = 2,k 2 = 4. Since they are real and different, the general solution has the form j (X)= C 1 e 2X + C 2 e 4x.

For case (b), the characteristic equation is the third-degree equation k 3 + 2k 2 - 3k = 0. Let's find the roots of this equation:

k(k 2 + 2 k - 3)= 0 Þ k = 0and k 2 + 2 k - 3 = 0 Þ k = 0, (k - 1)(k + 3) = 0,

T . e . k 1 = 0, k 2 = 1, k 3 = - 3.

These characteristic roots correspond to the fundamental system of solutions to the differential equation:

j 1 (X)= e 0X = 1, j 2 (X) = e x, j 3 (X)= e - 3X .

The general solution, according to formula (9), is the function

j (X)= C 1 + C 2 fx + C 3 e - 3X .

II ... All roots of the characteristic equation are different, but some of them are complex.

All coefficients of differential equation (14), and hence of its characteristic equation (15)- are real numbers, so if c among the characteristic roots there is a complex root k 1 = a + ib, that is, its conjugate root k 2 = ` k 1 = a- ib.To the first root k 1 corresponds to the solution of the differential equation (14)

j 1 (X)= e (a + ib)X = e a x e ibx = e ax(cosbx + isinbx)

(we used Euler's formula e i x = cosx + isinx). Similarly, the root k 2 = a- ib match solution

j 2 (X)= e (a - -ib)X = е a х е - ib x= e ax(cosbx - isinbx).

These solutions are complex. To obtain real solutions from them, we use the properties of solutions to a linear homogeneous equation (see 13.2). Functions

are real solutions to equation (14). Moreover, these solutions are linearly independent. Thus, the following conclusion can be drawn.

Rule 1.A pair of conjugate complex roots a± ib of the characteristic equation in the FSL of the linear homogeneous equation (14) matches two valid private solutionsand .

EXAMPLE 8. Find the general solution to the equation:

a) at¢ ¢ (X) - 2at ¢ (X) + 5at(X) = 0 ; b) at¢ ¢ ¢ (X) - at¢ ¢ (X) + 4at ¢ (X) - 4at(X) = 0.

Solution. In the case of equation (a), the roots of the characteristic equation k 2 - 2k + 5 = 0 are two conjugate complex numbers

k 1, 2 = .

Consequently, according to rule 1, they correspond to two real linearly independent solutions: and, and the general solution of the equation is the function

j (X)= C 1 e x cos 2x + C 2 e x sin 2x.

In case (b), to find the roots of the characteristic equation k 3 - k 2 + 4k- 4 = 0, factor its left side:

k 2 (k - 1) + 4(k - 1) = 0 Þ (k - 1)(k 2 + 4) = 0 Þ (k - 1) = 0, (k 2 + 4) = 0.

Therefore, we have three characteristic roots: k 1 = 1,k 2 , 3 = ± 2i. Cornu k 1 match solution , and a pair of conjugate complex roots k 2, 3 = ± 2i = 0 ± 2i- two valid solutions: and. We compose a general solution to the equation:

j (X)= C 1 fx + C 2 cos 2x + C 3 sin 2x.

III . There are multiples c among the roots of the characteristic equation.

Let k 1 - real root of multiplicity m characteristic equation (15), i.e., among the roots there is m equal roots. Each of them corresponds to the same solution to the differential equation (14) However, include m equal solutions in the FSR are impossible, since they constitute a linearly dependent system of functions.

It can be shown that in the case of a multiple root k 1 solutions to equation (14), in addition to the function, are the functions

The functions are linearly independent on the entire numerical axis, since, that is, they can be included in the FSR.

Rule 2. The real characteristic root k 1 multiplicity m in the FSR corresponds m solutions:

If k 1 - complex root of multiplicity m characteristic equation (15), then there is a conjugate root k 1 multiplicity m... By analogy, we obtain the following rule.

Rule 3. A pair of conjugate complex roots a± ib in FSR corresponds to 2m real linearly independent solutions:

, , ..., ,

, , ..., .

EXAMPLE 9. Find the general solution to the equation:

a) at¢ ¢ ¢ (X) + 3at¢ ¢ (X) + 3at¢ (X)+ at ( X) = 0; b) in IV(X) + 6at¢ ¢ (X) + 9at(X) = 0.

Solution. In case (a), the characteristic equation has the form

k 3 + 3 k 2 + 3 k + 1 = 0

(k + 1) 3 = 0,

i.e. k =- 1 - root of multiplicity 3. Based on rule 2, we write down the general solution:

j (X)= C 1 + C 2 x + C 3 x 2 .

The characteristic equation in case (b) is the equation

k 4 + 6k 2 + 9 = 0

or, otherwise,

(k 2 + 3) 2 = 0 Þ k 2 = - 3 Þ k 1, 2 = ± i.

We have a pair of conjugate complex roots, each of which has multiplicity 2. According to Rule 3, the general solution is written in the form

j (X)= C 1 + C 2 x + C 3 + C 4 x.

It follows from the above that for any linear homogeneous equation with constant coefficients, one can find a fundamental system of solutions and compose a general solution. Therefore, the solution of the corresponding inhomogeneous equation for any continuous function f(x) on the right-hand side can be found using the method of variation of arbitrary constants (see Section 5.3).

EXAMPLE 10 Using the variation method, find the general solution of the inhomogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = x e 2x .

Solution. First, we find the general solution of the corresponding homogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = 0. The roots of the characteristic equation k 2 - k- 6 = 0 are k 1 = 3,k 2 = - 2, a the general solution of the homogeneous equation - function ` at ( X) = C 1 e 3X + C 2 e - 2X .

We will seek a solution to the inhomogeneous equation in the form

at( X) = WITH 1 (X)e 3X + C 2 (X)e 2X . (*)

Find the Vronsky determinant

W[e 3X , e 2X ] = .

Let us compose the system of equations (12) for the derivatives of the unknown functions WITH ¢ 1 (X) and WITH¢ 2 (X):

Solving the system using Cramer's formulas, we obtain

Integrating, we find WITH 1 (X) and WITH 2 (X):

Substituting functions WITH 1 (X) and WITH 2 (X) into equality (*), we obtain a general solution of the equation at¢ ¢ (X) - at¢ (X) - 6at(X) = x e 2x :

In the case when the right-hand side of a linear inhomogeneous equation with constant coefficients has special view, a particular solution of the inhomogeneous equation can be found without resorting to the method of variation of arbitrary constants.

Consider the equation with constant coefficients

y (n) + a 1 y (n– 1) + ... a n – 1 y " + a n y = f (x), (16)

f( x) = eax(P n(x)cosbx + R m(x)sinbx), (17)

where P n(x) and R m(x) - polynomials of degree n and m respectively.

Private solution y *(X) of Eq. (16) is determined by the formula

at* (X) = x se ax(M r(x)cosbx + N r(x)sinbx), (18)

where M r(x) and N r(x) - polynomials of degree r = max(n, m) with undefined coefficients , a s equal to the multiplicity of the root k 0 = a + ib of the characteristic polynomial of Eq. (16), in this case it is assumed s = 0 if k 0 is not a characteristic root.

To compose a particular solution using formula (18), you need to find four parameters - a, b, r and s. The first three are determined by the right-hand side of the equation, and r- this is actually the highest degree x found on the right side. Parameter s is found by comparing the number k 0 = a + ib and the set of all (taking into account multiplicities) characteristic roots of equation (16), which are found by solving the corresponding homogeneous equation.

Consider special cases of the form of function (17):

1) at a ¹ 0, b= 0f(x)= e ax P n(x);

2) at a= 0, b ¹ 0f(x)= P n(x) Withosbx + R m(x)sinbx;

3) at a = 0, b = 0f(x)= P n(x).

Remark 1. If P n (x) º 0 or R m (x)º 0, then the right-hand side of the equation f (x) = e ax P n (x) with osbx or f (x) = e ax R m (x) sinbx, i.e., it contains only one of the functions - cosine or sine. But in the record of a particular solution, they must both be present, since, according to formula (18), each of them is multiplied by a polynomial with undefined coefficients of the same degree r = max (n, m).

EXAMPLE 11. Determine the form of a particular solution of a linear homogeneous equation of the 4th order with constant coefficients if the right-hand side of the equation is known f(X) = e x(2xcos 3x +(x 2 + 1)sin 3x) and the roots of the characteristic equation:

a ) k 1 = k 2 = 1, k 3 = 3,k 4 = - 1;

b ) k 1, 2 = 1 ± 3i,k 3, 4 = ± 1;

v ) k 1, 2 = 1 ± 3i,k 3, 4 = 1 ± 3i.

Solution. On the right-hand side, we find that in the particular solution at*(X), which is determined by formula (18), the parameters are: a= 1, b= 3, r = 2. They remain unchanged for all three cases, therefore, the number k 0, which defines the last parameter s formula (18) is equal to k 0 = 1+ 3i... In case (a), among the characteristic roots there is no number k 0 = 1 + 3i, means, s= 0, and a particular solution has the form

y *(X) = x 0 f x(M 2 (x)cos 3x + N 2 (x)sin 3x) =

= ex( (Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.

In case (b) the number k 0 = 1 + 3i occurs among the characteristic roots once, which means that s = 1 and

y *(X) = x e x((Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.

For case (c) we have s = 2 and

y *(X) = x 2 f x((Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.

In example 11 in the record of a particular solution there are two polynomials of the 2nd degree with undefined coefficients. To find a solution, you need to determine the numerical values ​​of these coefficients. Let's formulate a general rule.

To determine the unknown coefficients of the polynomials M r(x) and N r(x) equality (17) is differentiated the required number of times, the function y *(X) and its derivatives into equation (16). Comparing its left and right sides, one gets the system algebraic equations to find the coefficients.

EXAMPLE 12. Find a solution to the equation at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x, having determined the particular solution of the inhomogeneous equation by the form of the right-hand side.

Solution. The general solution to the inhomogeneous equation has the form

at( X) = ` at(X)+ y *(X),

where ` at ( X) - the general solution of the corresponding homogeneous equation, and y *(X) - particular solution of an inhomogeneous equation.

First, we solve the homogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = 0. Its characteristic equation k 2 - k- 6 = 0 has two roots k 1 = 3,k 2 = - 2, hence, ` at ( X) = C 1 e 3X + C 2 e - 2X .

We use formula (18) to determine the form of a particular solution at*(X). Function f(x) = xe 2x is a special case (a) of formula (17), while a = 2,b = 0 and r = 1, i.e. k 0 = 2 + 0i = 2. Comparing with characteristic roots, we conclude that s = 0. Substituting the values ​​of all parameters into formula (18), we have y *(X) = (Ah + B)e 2X .

To find the meanings A and V, find the derivatives of the first and second orders of the function y *(X) = (Ah + B)e 2X :

y *¢ (X)= Ae 2X + 2(Ah + B)e 2X = (2Ax + A + 2B)e 2x,

y *¢ ¢ (X) = 2Ae 2X + 2(2Ax + A + 2B)e 2X = (4Ah + 4A + 4B)e 2X .

After function substitution y *(X) and its derivatives into the equation, we have

(4Ah + 4A + 4B)e 2X - (2Ax + A + 2B)e 2X - 6(Ah + B)e 2X = xe 2x Þ Þ A =- 1/4,B =- 3/16.

Thus, a particular solution of the inhomogeneous equation has the form

y *(X) = (- 1/4X- 3/16)e 2X ,

and the general solution - at ( X) = C 1 e 3X + C 2 e - 2X + (- 1/4X- 3/16)e 2X .

Remark 2.In the case when the Cauchy problem is posed for an inhomogeneous equation, one must first find the general solution of the equation

at( X) = ,

having determined all the numerical values ​​of the coefficients in at*(X). Then use the initial conditions and, substituting them into the general solution (and not into y *(X)), find the values ​​of the constants C i.

EXAMPLE 13. Find a solution to the Cauchy problem:

at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x , at(0) = 0, at ¢ (X) = 0.

Solution. The general solution to this equation

at(X) = C 1 e 3X + C 2 e - 2X + (- 1/4X- 3/16)e 2X

was found in Example 12. To find a particular solution satisfying the initial conditions of this Cauchy problem, we obtain the system of equations

Solving it, we have C 1 = 1/8, C 2 = 1/16. Therefore, the solution to the Cauchy problem is the function

at(X) = 1/8e 3X + 1/16e - 2X + (- 1/4X- 3/16)e 2X .

Remark 3(superposition principle). If in the linear equation L n[y(x)]= f(x), where f(x) = f 1 (x)+ f 2 (x) and y * 1 (x) - equation solution L n[y(x)]= f 1 (x), a y * 2 (x) - equation solution L n[y(x)]= f 2 (x), then the function y *(X)= y * 1 (x)+ y * 2 (x) is an by solving the equation L n[y(x)]= f(x).

PRI me R14. Indicate the form of the general solution of a linear equation

at¢ ¢ (X) + 4at(X) = x + sinx.

Solution. General solution of the corresponding homogeneous equation

` at(x) = C 1 cos 2x + C 2 sin 2x,

since the characteristic equation k 2 + 4 = 0 has roots k 1, 2 = ± 2i.Right part equation does not correspond to formula (17), but if we introduce the notation f 1 (x) = x, f 2 (x) = sinx and use the principle of superposition , then a particular solution of the inhomogeneous equation can be found in the form y *(X)= y * 1 (x)+ y * 2 (x), where y * 1 (x) - equation solution at¢ ¢ (X) + 4at(X) = x, a y * 2 (x) - equation solution at¢ ¢ (X) + 4at(X) = sinx. According to the formula (18)

y * 1 (x) = Ax + B,y * 2 (x) = Ссosx + Dsinx.

Then the particular solution

y *(X) = Ax + B + Csosx + Dsinx,

therefore, the general solution has the form

at(X) = C 1 cos 2x + C 2 e - 2X + A x + B + Ccosx + Dsinx.

PRI me R15. The electrical circuit consists of a series-connected current source with emf e(t) = E sinw t, inductance L and capacity WITH, and

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

Consider a second-order linear differential equation (LDDE) with constant coefficients of the form y "" + p x.

Let us turn 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

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 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 LODE and some particular solution y ~, where the original inhomogeneous equation is y = y 0 + y ~.

Hence it is seen 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. Then you should go to the definition of y ~.

The choice of a particular solution to the LNDE depends on the form of the existing function f (x) located on the right side of the equation. For 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 degree n f (x) = P n (x), it follows that a particular solution of the LNDE 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 y ~ is a particular solution y ~ "" + p y ~ "+ q y ~ = f (x), then the available coefficients, which are determined by the polynomial
Q n (x), we find using the method of undefined coefficients from the equality y ~ "" + p · y ~ "+ q · y ~ = f (x).

Example 1

Calculate by 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, which corresponds to the equation y 0 or a particular solution to the inhomogeneous equation y ~, that is, y = y 0 + y ~.

To begin with, we will 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 you find the roots. We get that

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

Got the roots different and valid. Therefore, we write

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

Find y ~. It can be seen that the right side given equation is a polynomial of the second degree, then one of the roots is equal to zero. From this we obtain that the 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 undefined coefficients.

Let us 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 find the coefficients and write down: 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 notation is called the general solution of the original linear inhomogeneous differential equation of the second order 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 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 xx = 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 exponent f (x) = P n (x) eax, then we obtain from this that an equation of the form y ~ = eax Q n ( x) x γ, where Q n (x) is a polynomial of degree n, 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 specified equation corresponds to the LODE y "" - 2 y "= 0. From the previous example, you can see that its roots are equal to k 1 = 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, the 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 has no 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 ~ "= ex A x 2 + B x + C" = ex A x 2 + B x + C + ex 2 A x + B = = ex A x 2 + x 2 A + B + B + C y ~ "" = ex A x 2 + x 2 A + B + B + C "= = ex A x 2 + x 2 A + B + B + C + ex 2 A x + 2 A + B = = ex A x 2 + x 4 A + B + 2 A + 2 B + C

y ~ "" - 2 y ~ "= (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 = x 2 + 1 ex ⇔ ex - A x 2 - B x + 2 A - C = (x 2 + 1) ex ⇔ - 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 get the system 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 ~ = ex (A x 2 + B x + C) = ex - x 2 + 0 x - 3 = - ex x 2 + 3 is a particular solution of the LNDE, and y = y 0 + y = C 1 e 2 x - ex · x 2 + 3 - the general solution for the inhomogeneous second-order differential 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 an equation of the form y ~ = A cos β x + B sin β x x γ, where A and B are considered as indefinite coefficients, and r as the number of complex conjugate roots related to the characteristic equation, equal to ± i β ... In this case, the search for the coefficients is carried out according to 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 = 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 from the characteristic equation are considered to be the conjugate pair ± 2 i, then f (x) = cos (2 x) + 3 sin (2 x). Hence it is clear that the search for y ~ will be performed from y ~ = (A cos (β x) + B sin (β x) x γ = (A cos (2 x) + B sin (2 x)) x. Unknown the 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 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 the 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 LDE with constant coefficients is

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) = eax P n (x) sin (β x) + Q k (x) cos (β x), then y ~ = eax (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, m, where m = m a x (n, k)... Finding the coefficients L m (x) and N m (x) is produced proceeding from 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

By 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, having previously written down 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

Got the roots to be valid and distinct. Hence y 0 = C 1 e x + C 2 e 2 x. Next, it is necessary to seek 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 obtained 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

It follows from everything 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 the general solution of the given linear equation is 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 observance of 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 particular solutions of the LODE, C 1 and C 2 are considered arbitrary constants;
• adoption as a general decision of the LNDE y = C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2;
• definition of the 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 the functions C 1 (x) and C 2 (x) by means of 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 down y 0, y "" + 36 y = 0. Let's write down 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 receive the form y = C 1 (x) · cos (6 x) + C 2 (x) · sin (6 x). It is necessary to go to the definition of the derivatives of functions C 1 (x) and C 2 (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)) = = 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 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 should 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

From this it follows that the general solution will be as follows:

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

The 2nd order LNDE 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.

In relation to LNDU 2 with PC, the following two statements are true.

Suppose that some function $ U $ is an arbitrary particular solution of an inhomogeneous differential equation. Suppose also that some function $ Y $ is a general solution (GR) of the corresponding linear homogeneous differential equation (LDE) $ y "" + p \ cdot y "+ q \ cdot y = 0 $. Then the GD of LDE-2 is equal to the sum of the indicated particular and general solutions, that is, $ y = U + Y $.

If the right side of the 2nd order LNDE is a 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) $, which correspond to each of the functions $ f_ ( 1) \ left (x \ right), f_ (2) \ left (x \ right), ..., f_ (r) \ left (x \ right) $, and only after that write the CR LNDE-2 in the form $ U = U_ (1) + U_ (2) + ... + U_ (r) $.

2nd order LNDU solution from PC

Obviously, the form of this or that 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 LNDE-2 are formulated in the form of the following four rules.

Rule number 1.

The right side of LNDU-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 PD $ 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 that 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 zero... The coefficients of the polynomial $ Q_ (n) \ left (x \ right) $ are found by the method of undefined coefficients (NK).

Rule number 2.

The right side of LNDU-2 is $ 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 side of LNDU-2 is $ 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 sought 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 NK method.

Rule number 4.

The right side of the LNDE-2 is $ 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 sought 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 NDT 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 of the LNDE-2, it is necessary:

• substitute the PD $ U $ written in general view, to the left side of LNDU-2;
• on the left side of LNDU-2, simplify and group members with the same powers of $ x $;
• in the resulting identity, equate the coefficients of the terms with the same powers of $ x $ of the left and right sides;
• solve the resulting system of linear equations for unknown coefficients.

Example 1

Problem: find OR LNDU-2 $ y "" - 3 \ cdot y "-18 \ cdot y = \ left (36 \ cdot x + 12 \ right) \ cdot e ^ (3 \ cdot x) $. Find also PD satisfying the initial conditions $ y = 6 $ for $ x = 0 $ and $ y "= 1 $ for $ x = 0 $.

We write down the corresponding LODU-2: $ y "" - 3 \ cdot y "-18 \ cdot y = 0 $.

Characteristic equation: $ k ^ (2) -3 \ cdot k-18 = 0 $. Roots of the characteristic equation: $ k_ (1) = -3 $, $ k_ (2) = 6 $. These roots are valid and different. 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 side of this LNDU-2 is $ \ left (36 \ cdot x + 12 \ right) \ cdot e ^ (3 \ cdot x) $. In it, 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 PD of this LNDE-2 has the form $ U = \ left (A \ cdot x + B \ right) \ cdot e ^ (3 \ cdot x) $.

We will search for the coefficients $ A $, $ B $ by the NK method.

Find the first PD derivative:

$ 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 PD:

$ 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). $

Substitute the functions $ U "" $, $ U "$ and $ U $ instead of $ y" "$, $ y" $ and $ y $ into the given LNDU-2 $ y "" - 3 \ cdot y "-18 \ cdot y = \ left (36 \ cdot x + 12 \ right) \ cdot e ^ (3 \ cdot x). $ In this case, since the exponent $ e ^ (3 \ cdot x) $ enters 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 the 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 apply the NDT 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 as follows: $ A = -2 $, $ B = -1 $.

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) $.

OP $ 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 satisfying 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). $

Substitute in $ y $ and $ y "$ the initial conditions $ y = 6 $ at $ x = 0 $ and $ y" = 1 $ at $ 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) $ by 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) $.