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

Educational institution "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Guidelines

to study the topic “Linear differential equations of the second order” by students of the accounting faculty 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 called an equation of the form

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

If
on the interval
, then equation (1) will take the form

, (2)

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

Consider the complex function

, (3)

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

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

If And there are solutions to 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 such numbers exist And
, not equal to zero at the same time, that on this interval the equality

If equality (4) occurs only when
And
, then the functions
And
are called linearly independent on the interval
.

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

Example 2 . Functions
And
are linearly independent on any interval, since the equality
is possible only in the case when
, And
.

Construction of a general solution to a linear homogeneous

equations

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

We will look for linearly independent solutions to equation (2) in the form

, (5)

Where – a certain number. Then
,
. Let's substitute these expressions into equation (2):

or
.

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

Let the roots And characteristic equations are real and distinct. Then the solutions to equation (2) will be the functions
And
. These solutions are linearly independent, since the equality
can only be carried out when
, And
. Therefore, the general solution to 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 the differential equation. The general solution to this equation is
.

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

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

Example 4 . Solve quadratic equation
.

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

Let the roots of the characteristic equation be complex, i.e.
,
, Where
. Solutions of 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 to this equation are both the real and imaginary parts of this function. Thus, the solutions to equation (2) will be the functions
And
. Since equality

can only be executed if
And
, then these solutions are linearly independent. Therefore, 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 of 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 is:

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

Inhomogeneous linear differential equations of the second order with constant coefficients

and the 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). Let's look at cases where this is possible.

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

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

If
is the root of the characteristic equation, then a particular solution to 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
. Its characteristic equation
has roots
And
. The general solution of the homogeneous equation has the form
.

Because
is not a root of the characteristic equation, then we will look for a particular solution of the inhomogeneous equation in the form of a function
. Let's find the derivatives of this function
,
and substitute them into this equation:

or . Let us equate the coefficients for and free members:
Having solved this system, we get
,
. Then a particular solution of the inhomogeneous equation has the form
, and the general solution of a given 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 to the inhomogeneous equation should be sought in the form. If
is the root of the characteristic multiplicity equation k (k=1 or k=2), then in this case a 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, a particular solution to the inhomogeneous equation should be sought in the form
. Let's find the derivatives of the first and second orders:

Let's substitute into the differential equation:
+ +,
+,.

Let us equate the coefficients for and free members:

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

.

Lagrange method of variation of arbitrary constants

The method of varying arbitrary constants can be applied to any inhomogeneous linear equation with constant coefficients, regardless of the type of right-hand side. This method allows you to always find a general solution to an inhomogeneous equation if the general solution to 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 varying arbitrary constants is that the general solution to equation (1) is sought in the form

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

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

And
.

Substituting these expressions into (9), we obtain a general solution to 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 a given differential equation is
. Its roots are complex
,
. Because
And
, That
,
, and the general solution of the homogeneous equation has the form. Then we will look for a general solution to this inhomogeneous equation in the form where
And
- unknown functions.

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

Having solved this system, we find
,
. Then

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

This is the general solution to this differential equation, obtained using 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 to a linear inhomogeneous equation sought if the roots of the characteristic equation are different and not equal to zero, and the right side of the equation is a polynomial of degree m?

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

What is the essence of Lagrange's method?

We have seen that, in the case where the general solution of a linear homogeneous equation is known, it is possible to find the general solution of an inhomogeneous equation using the method of variation of arbitrary constants. However, the question of how to find a general solution to a homogeneous equation remained open. In the special case when in the linear differential equation (3) all 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 look for a solution to equation (14) in the form j (X) = e kx.

Let us substitute the function into equation (14) j (X) and its order derivatives m (1 £ m£ n)j (m) (X) = k m e 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 side- characteristic polynomial , its roots- characteristic roots differential equation (14).

Conclusion:

functionj (X) = e kx - solution to linear homogeneous equation (14) if and only if the number k - 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 distinct.

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

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

Where WITH 1 , C 2 , ..., C n - arbitrary constants.

Example 7. Find the general solution of the linear homogeneous equation:

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

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

k(at (m) (x) « k m),

while the function itself at(X) as the zeroth 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 3rd 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 = 0i 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 of 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 e x + C 3 e - 3X .

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

All coefficients of the differential equation (14), and therefore of its characteristic equation (15)- real numbers, which means 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). Likewise, the root k 2 = a- ib corresponds to the solution

j 2 (X)= e (a - -ib)X = e a x e - 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 of equation (14). Moreover, these solutions are linearly independent. Thus, we can draw the following conclusion.

Rule 1.A pair of conjugate complex roots a± ib of the characteristic equation in the FSR of the linear homogeneous equation (14) corresponds to two real partial 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, we factorize 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 corresponds to the 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 e x + C 2 cos 2x + C 3 sin 2x.

III . Among the roots of the characteristic equation there are multiples.

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 There are no equal solutions in the FSR, 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. Real characteristic root k 1 multiplicity m in the FSR corresponds m solutions:

If k 1 - complex root 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 the FSR corresponds to 2mreal linearly independent solutions:

, , ..., ,

, , ..., .

Example 9. Find the general solution to the equation:

A) at¢ ¢ ¢ (X) + 3at¢ ¢ (X) + 3at¢ (X)+ y ( X)= 0;b) at 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 as

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

From the above it follows that for any linear homogeneous equation with constant coefficients it is possible to find a fundamental system of solutions and compose a general solution. Consequently, the solution to the corresponding inhomogeneous equation for any continuous function f(x) on the right 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 to the inhomogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x .

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

We will look for a solution to the inhomogeneous equation in the form

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

Let's find the Wronski determinant

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

Let us compose a system of equations (12) for the derivatives of 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 to the equation at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x :

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

Consider the equation with constant coefficients

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

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

Where P n(x) And R m(x) - degree polynomials n And m respectively.

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

at* (X) = xse ax(Mr(x)cosbx + Nr(x)sinbx), (18)

Where Mr(x) And Nr(x) - degree polynomials r = max(n, m) with uncertain coefficients , A s equal to the multiple of the root k 0 = a + ib characteristic polynomial of equation (16), and we assume s = 0 if k 0 is not a characteristic root.

In order to compose a particular solution using formula (18), you need to find four parameters - a, b, r And s. The first three are determined from the right side of the equation, and r- this is actually the highest degree x, found on the right side. Parameter s found from comparison of numbers 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.

Let us consider special cases of the form of function (17):

1) at a ¹ 0, b= 0f(x)= e ax P n(x);

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

3) when a = 0, b = 0f(x)=Pn(x).

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

Example 11. Determine the type of partial solution to a linear homogeneous equation of the 4th order with constant coefficients if the right 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 side we find that in the particular solution at*(X), which is determined by formula (18), parameters: a= 1, b= 3, r = 2. They remain the same for all three cases, hence the number k 0 which specifies the last parameter s formula (18) is equal to k 0 = 1+ 3i. In case (a) there is no number among the characteristic roots k 0 = 1 + 3i, Means, s= 0, and a particular solution has the form

y*(X) = x 0 e 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 once among the characteristic roots, which means 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 e x((Ax 2 +Bx+C)cos 3x+(A 1 x 2 +B 1 x+C 1)sin 3x.

In example 11, the particular solution contains two polynomials of degree 2 with undetermined coefficients. To find a solution, you need to determine the numerical values ​​of these coefficients. Let us formulate a general rule.

To determine the unknown coefficients of polynomials Mr(x) And Nr(x) equality (17) is differentiated the required number of times, and the function is substituted y*(X) and its derivatives into equation (16). By comparing its left and right sides, a system of algebraic equations is obtained for finding the coefficients.

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

Solution. The general solution of 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 a non-homogeneous 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 .

Let us use formula (18) to determine the type of particular solution at*(X). Function f(x) = xe 2x represents a special case (a) of formula (17), while a = 2,b = 0 And r = 1, i.e. k 0 = 2 + 0i = 2. Comparing with the 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 values A And IN, let's find the first and second order derivatives of the function y*(X) = (Ah + B)e 2X :

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

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

Thus, a particular solution to 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 .

Note 2.In the case when the Cauchy problem is posed for an inhomogeneous equation, one must first find a general solution to 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 ,y(0) = 0, y ¢ (X) = 0.

Solution. The general solution to this equation is

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 that satisfies the initial conditions of this Cauchy problem, we obtain a 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 .

Note 3(superposition principle). If in a linear equation Ln[y(x)]= f(x), Where f(x) = f 1 (x)+ f 2 (x) And y* 1 (x) - solution to the equation Ln[y(x)]= f 1 (x), A y* 2 (x) - solution to the equation Ln[y(x)]= f 2 (x), then the function y*(X)= y* 1 (x)+ y* 2 (x) is solving the equation Ln[y(x)]= f(x).

Example 14. Indicate the type of general solution to 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.The right side of the 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 to the inhomogeneous equation can be found in the form y*(X)= y* 1 (x)+ y* 2 (x), Where y* 1 (x) - solution to the equation at¢ ¢ (X) + 4at(X) = x, A y* 2 (x) - solution to the equation at¢ ¢ (X) + 4at(X) = sinx. According to formula (18)

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

Then the particular solution

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

therefore, the general solution has the form

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

Example 15. An electrical circuit consists of a current source connected in series with an emf e(t) = E sinw t, inductance L and containers WITH, and

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

A 2nd order LDDE 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.

With regard to LNDU 2 with PC, the following two statements are true.

Let us assume that some function $U$ is an arbitrary partial solution of an inhomogeneous differential equation. Let us also assume that some function $Y$ is the general solution (GS) of the corresponding linear homogeneous differential equation (HLDE) $y""+p\cdot y"+q\cdot y=0$. Then the GR of LHDE-2 is equal to the sum of the indicated private and general solutions, that is, $y=U+Y$.

If the right-hand side of a 2nd order LMDE 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 we can find the PDs $U_(1) ,U_(2) ,...,U_(r)$ 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 CR LNDU-2 in the form $U=U_(1) +U_(2) +...+U_(r) $.

Solution of 2nd order LPDE with PC

It is obvious that the type of one or another PD $U$ of a given LNDU-2 depends on the specific form of its right-hand side $f\left(x\right)$. The simplest cases of searching for PD LNDU-2 are formulated in the form of the following four rules.

Rule #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 that are equal to zero. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the method of indefinite coefficients (UK).

Rule No. 2.

The right side of LNDU-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 NC method.

Rule No. 3.

The right side of LNDU-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 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 using the non-destructive method.

Rule No. 4.

The right side of LNDU-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 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 NC method.

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

• substitute the PD $U$, written in general form, into the left side of LNDU-2;
• on the left side of LNDU-2, perform simplifications and group terms with the same powers $x$;
• in the resulting identity, equate the coefficients of 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 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 LOD-2: $y""-3\cdot y"-18\cdot y=0$.

Characteristic equation: $k^(2) -3\cdot k-18=0$. The roots of the characteristic equation are: $k_(1) =-3$, $k_(2) =6$. These roots are valid 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 side of this LNDU-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 $\alpha =3$. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PD of this LNDU-2 has the form $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $.

We will search for the coefficients $A$, $B$ using the NC method.

We find the first derivative of the Czech Republic:

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

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

We find the second derivative of the Czech Republic:

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

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

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

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

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 use the NDT method. We obtain 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: $A=-2$, $B=-1$.

PD $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 $y"$ of the OP:

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

We substitute into $y$ 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 received a system of equations:

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

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

Let's solve it. We find $C_(1) $ using Cramer's formula, and $C_(2) $ we determine 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 has the form: $y=4\cdot e^(-3\cdot x) +3\cdot e^(6\cdot x) +\left(-2\cdot x-1\right )\cdot e^(3\cdot x) $.

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 a 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 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 ~.

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 of the 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 general equation is y = y 0 + y ~ . The indicated equation corresponds to the 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 can be seen that the right side of the 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 convert:

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;
• adoption as a general solution of 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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