Linear equations with constant coefficients online. Linear inhomogeneous second-order differential equations with constant coefficients
We were convinced that, in the case when it is known common decision Linear uniform equation, It is possible by the method of variation of arbitrary constants to find the general solution of the inhomogeneous equation. However, the question of how to find a general solution of a homogeneous equation remained open. In the particular case, when in the linear differential equation (3) all coefficients r I.(h.) \u003d A I. - Constants, it is solved quite simply, even without integration.
Considerable uniform differential equation with permanent coefficients, i.e. equations of the form
y. (n.) + A. 1 y. (n. – 1) + ... A. N. – 1 y. " + a n y \u003d 0, (14)
where a I.- Constants (i.= 1, 2, ..., N.).
As is known, for a linear homogeneous 1st order equation, the solution is the function e. KX.We will look for the solution of equation (14) in the form of j. (h.) = e. KX..
Substitute to equation (14) function j. (h.) and its derivative order m. (1 £ M.£ N.)j. (m.) (h.) = k M E KX. Receive
(k n + a 1 k N. – 1 + ... and n – 1 k + A N)e KX \u003d.0,
but e. K H. ¹ 0 at any H., so
k n + and 1 k n – 1 + ... A. N. – 1 k + and n \u003d0. (15)
Equation (15) is called characteristic equation a polynomial standing in the left side- characteristic polynomial His roots- characteristic roots Differential equation (14).
Output:
functionj. (h.) = e. KX. - solving a linear homogeneous equation (14) if and only when the number k. - The root of the characteristic equation (15).
Thus, the process of solving a linear homogeneous equation (14) reduces to the solution of an algebraic equation (15).
Various cases of characteristic roots are possible.
1. All the roots of the characteristic equation are valid and different.
In this case n. Various characteristic roots K. 1 , K. 2 , ..., k n correspond to n. Different solutions of a homogeneous equation (14)
It can be shown that these solutions are linearly independent, therefore form fundamental system solutions. Thus, the general solution of the equation is a function
where FROM 1 , C. 2 , ..., with N - Arbitrary constants.
PRI ME R7. Find a general solution of a linear homogeneous equation:
but) w.¢ ¢ (h.) - 6w.¢ (h.) + 8w.(h.) \u003d 0, b) w.¢ ¢ ¢ (h.) + 2w.¢ ¢ (h.) - 3w.¢ (h.) = 0.
Decision. Let's make a characteristic equation. To do this, replace the derivative order m. Functions y.(x.) on the relevant degree
k.(w. (m.) (x.) « k M.),
in this case, the function itself w.(h.) as a zero order derivative is replaced by k. 0 = 1.
In the case of (a), the characteristic equation has the form k. 2 - 6k +.8 = 0. Roots of this square equation k. 1 = 2, K. 2 = 4. Since they are valid and different, the general solution has the form j. (h.) \u003d S. 1 e. 2h. + With 2 e. 4x
For the case (b), the characteristic equation is 3-ystepy equation k. 3 + 2k. 2 - 3k \u003d. 0. We will 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.
This characteristic root corresponds to the fundamental system of solutions of the differential equation:
j. 1 (h.) \u003d E. 0h. = 1, j. 2 (h.) \u003d E H., j. 3 (h.) \u003d E. - 3h. .
General decision, according to formula (9), is a function
j. (h.) \u003d S. 1 + S. 2 e x + with 3 e. - 3h. .
II. . All the roots of the characteristic equation are different, but among them there are complex.
All coefficients of the differential equation (14), and therefore its characteristic equation (15)- Actual numbers, which means, if C had characteristic roots there is a complex root k. 1 \u003d a + ib,that is, the root conjugate k. 2 = ` k. 1 \u003d A.- IB.First root k. 1 corresponds to the solution of the differential equation (14)
j. 1 (h.) \u003d E. (a + IB.)h. \u003d E A X E IBX \u003d E Ah(cosbx + Isinbx.)
(used the formula Euler e i x \u003d cosx + isinx). Similarly, Korni. k. 2 \u003d A.- IB.corresponds to the decision
j. 2 (h.) \u003d E. (a - -B.)h. \u003d e a x e - IB H.\u003d E Ah.(cosbx - isinbx).
These solutions are complex. To obtain actual solutions from them, we use the properties of solutions of a linear homogeneous equation (see 13.2). Functions
are valid solutions equation (14). In addition, these solutions are linearly independent. Thus, you can draw the following conclusion.
Rule 1.. Pair of conjugate complex roots± Ib characteristic equation in the FSR linear homogeneous equation (14) corresponds to two valid private solutionsand .
PRI MERE P8. Find a general solution of the equation:
but) w.¢ ¢ (h.) - 2w. ¢ (h.) + 5w.(h.) = 0 ; b) w.¢ ¢ ¢ (h.) - W.¢ ¢ (h.) + 4w. ¢ (h.) - 4w.(h.) = 0.
Decision. In the case of equation (A) the roots of the characteristic equation K. 2 - 2k +.5 \u003d 0 are two conjugate integrated numbers
k. 1, 2 = .
Consequently, it, according to Rule 1, corresponds to two valid linear independent solutions: and, and the general solution of the equation is the function
j. (h.) \u003d S. 1 E X Cos.2x + S. 2 E X SIN2x.
In the case (b) to find the roots of the characteristic equation k. 3 - K. 2 + 4k.- 4 = 0, spread its left part on the factors:
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 root: k. 1 = 1, K 2. , 3 = ± 2i.Korni. k. 1 Corresponds to the decision , and a pair of conjugate complex roots K. 2, 3 = ± 2i \u003d.0 ± 2i.- two valid solutions: and. We compile a general solution of the equation:
j. (h.) \u003d S. 1 e x + with 2 cos.2x + S. 3 sin.2x.
III . C Redi the roots of the characteristic equation are multiple.
Let be k. 1 - Valid root of multiplicity M.characteristic equation (15), i.e. among the roots there m. equal roots. Each of them corresponds to the same solution of the differential equation (14) however, M. Equal solutions in the FSR cannot, since they constitute a linearly dependent system of functions.
It can be shown that in the case of a multiple root K 1. solutions equation (14), in addition to function are functions
Functions are linearly independent on the entire numeric axis, since, i.e. they can be included in the FSR.
Rule 2. Valid characteristic root k. 1 multiplicity m.in the FSR corresponds m. Solutions:
If a k. 1 - Complex root of multiplicity M.characteristic equation (15), then there is a conjugated root k. 1 multiplicity M.. By analogy, we obtain the following rule.
Rule 3.. Pair of conjugate complex roots a± IB in the FSER corresponds to 2 media linearly independent solutions:
, , ..., ,
, , ..., .
PRI MERE P9. Find a general solution of the equation:
but) w.¢ ¢ ¢ (h.) + 3w.¢ ¢ (h.) + 3w.¢ (h.) + U. ( h.) \u003d 0; b) iV(h.) + 6w.¢ ¢ (h.) + 9w.(h.) = 0.
Decision. In the case of (a), the characteristic equation has the form
k. 3 + 3 k. 2 + 3 k. + 1 = 0
(k +.1) 3 = 0,
i.e. k \u003d.- 1 - Root of multiplicity 3. On the basis of rule 2, write a general solution:
j. (h.) \u003d S. 1 + S. 2 x + S. 3 X. 2 .
The characteristic equation in the case of (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 is multiplicity 2. According to Rule 3, the general solution is written in the form of
j. (h.) \u003d S. 1 + S. 2 x + S. 3 + S. 4 x.
From above, it follows that for any linear homogeneous equation with constant coefficients, you can find a fundamental system of solutions and make a general solution. Consequently, the solution of the corresponding inhomogeneous equation for any continuous function f.(x.) You can find on the right side using the method of variation of arbitrary constants (see paragraph 5.3).
PRI ME R10.MEMEMENT Variations to find a general solution of an inhomogeneous equation w.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) = x E. 2x. .
Decision. We will first find a general solution of the corresponding homogeneous equation. w.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) \u003d 0. Roots of the characteristic equation k. 2 - K.- 6 = 0 are k. 1 = 3, K. 2 = - 2, A. general solution of a homogeneous equation - function ` w. ( h.) \u003d S. 1 e. 3h. + S. 2 E. - 2h. .
We will look for the solution of an inhomogeneous equation in the form
w.( h.) = FROM 1 (h.)e. 3h. + S. 2 (h.)e. 2h. . (*)
We find the determinant of Vronsky
W.[e. 3h. , E. 2h. ] = .
We will make a system of equations (12) regarding derivative unknown functions FROM ¢ 1 (h.) I. FROM¢ 2 (h.):
Solving the system by crawler formulas, we get
Integrating, we find FROM 1 (h.) I. FROM 2 (h.):
Substituting functions FROM 1 (h.) I. FROM 2 (h.) In equality (*), we obtain the general solution of the equation w.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) = x E. 2x. :
In case when right part linear inhomogeneous equation with constant coefficients has special view, the particular solution of the inhomogeneous equation can be found without resorting to the method of variation of arbitrary constants.
Consider equation with permanent coefficients
y. (n.) + a 1 y (n.– 1) + ... A. N. – 1 y. " + a n y \u003d f (x.), (16)
f.( x.) = e. AX.(P N.(x.)cOSBX + R M(x.)sinbx), (17)
where P N.(x.) I. R M.(x.) - polynomials n. and m. respectively.
Private solution y *(h.) equations (16) is determined by the formula
w.* (h.) = x S.e. AX.(M R.(x.)cOSBX + N R(x.)sinbx), (18)
where M R.(x.) and N R.(x.) - polynomials r \u003d Max(n, M.) with uncertain coefficients , but s. Equally the multiplicity of the root k. 0 \u003d A + IB characteristic polynomial equation (16), while supposed s \u003d.0, if K. 0 is not a characteristic root.
To make a special decision according to the formula (18), you need to find four parameters - a, B, R and s.The first three are determined by the right part of the equation, and r.- it is actually the highest degree x.found in the right part. Parameter S. is from the comparison of the number k. 0 \u003d A + IB and a set of all (in mindcutivity) of the characteristic roots of equation equation (16), which are in solving the corresponding homogeneous equation.
Consider special cases of the type of function (17):
1) for a. ¹ 0, b.= 0f.(x.)= e AX P N(x.);
2) for a.= 0, b. ¹ 0f.(x.)= P N.(x.) fromoSBX + R M(x.)sinbx;
3) for a. = 0, b. = 0f.(x.) \u003d P n.(x.).
Remark 1. If P n (x) º 0 or R M (X)º 0, then the right side of the equation (x) \u003d e ax p n (x) with OSBX or F (X) \u003d E AX R M (x) sinbx, i.e. it contains only one of the functions - Cosine or sine. But in the record of a private solution, they must be present both, since, according to formula (18), each of them is multiplied by a polynomial with uncertain coefficients of the same degree R \u003d MAX (N, M).
PRI M E P 11. Determine the type of particular solution of the linear homogeneous 4th order equation with constant coefficients, if the right part of the equation is known f.(h.) \u003d E X.(2xcos.3x +.(x. 2 + 1)sin.3x.) and the roots of the characteristic equation:
but ) k. 1 \u003d K. 2 = 1, k. 3 = 3, K. 4 = - 1;
b. ) k. 1, 2 = 1 ± 3i., K. 3, 4 = ± 1;
in ) k. 1, 2 = 1 ± 3i., K. 3, 4 = 1 ± 3i.
Decision. On the right, we find that in a private solution w.*(h.), which is determined by the formula (18), parameters: a.= 1, b.= 3, r \u003d.2. They remain unchanged for all three cases, therefore, the number K. 0, which defines the last parameter s. Formulas (18) is equal K. 0 = 1+ 3i.. In the case of (a) among the characteristic roots there is no number K. 0 = 1 + 3i,it means s.\u003d 0, and the private solution has the form
y *(h.) = x. 0 e X.(M. 2 (x.)cos.3x + N. 2 (x.)sin.3x.) =
= e. X.( (AX. 2 + BX + C)cos.3x +.(A. 1 x. 2 + B. 1 x + C. 1)sin.3x.
In the case (b) number k. 0 = 1 + 3i.it occurs among the characteristic roots once, it means that S \u003d.1 and
y *(h.) \u003d x e x((AX. 2 + BX + C)cos.3x +.(A. 1 x. 2 + B. 1 x + C. 1)sin.3x.
For the case (c) we have s \u003d.2 I.
y *(h.) \u003d H. 2 E X.((AX. 2 + BX + C)cos.3x +.(A 1. x. 2 + B. 1 x + C. 1)sin.3x.
In Example 11, there are two polynomials of a 2nd degree with uncertain coefficients in the recording of a private solution. To find a solution, it is necessary to determine the numeric values \u200b\u200bof these coefficients. We formulate a general rule.
To determine unknown polynomial coefficients M R.(x.) I. N R.(x.) Equality (17) differentiate the desired number once, substitute the function y *(h.) and its derivatives in equation (16). Comparing its left and right parts, receive the system algebraic equations To find the coefficients.
PRI ME R 12. Find a solution to the equation w.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) = xe. 2x. , Identify the private solution of the inhomogeneous equation from the right side.
Decision. The general solution of the inhomogeneous equation has the form
w.( h.) = ` w.(h.) + *(h.),
where ` w. ( h.) - general solution of the corresponding homogeneous equation, and y *(h.) - Private solution of an inhomogeneous equation.
First solve a homogeneous equation W.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) \u003d 0. Its characteristic equation k. 2 - K.- 6 = 0 it has two roots k. 1 = 3, K. 2 = - 2, hence, ` w. ( h.) \u003d S. 1 e. 3h. + S. 2 e. - 2h. .
We use the formula (18) to determine the type of private solution w.*(h.). Function f.(x.) = xe. 2x. represents a special case (a) of formula (17), while a \u003d.2, B \u003d.0 and r \u003d.1, i.e. k. 0 = 2 + 0i \u003d.2. Comparing with characteristic roots, conclude that s \u003d.0. Substituting the values \u200b\u200bof all parameters in formula (18), we have y *(h.) = (Ah + B.)e. 2h. .
To find values BUT and IN, find the derivatives of the first and second orders of function y *(h.) = (Ah + B.)e. 2h. :
y *¢ (h.) \u003d Ae. 2h. + 2(Ah + B.)e. 2h. = (2Ah + a +2B.)e. 2x
y *¢ ¢ (h.) = 2Ae 2h. + 2(2Ah + a +2B.)e. 2h. = (4Ah +.4A +.4B.)e. 2h. .
After the functions of the function y *(h.) and its derivatives in the equation have
(4Ah +.4A +.4B.)e. 2h. - (2Ah + a +2B.)e. 2h. - 6(Ah + B.)e. 2h. \u003d Xe. 2x. Þ Þ A \u003d.- 1/4, B \u003d.- 3/16.
Thus, the particular solution of the inhomogeneous equation has the form
y *(h.) = (- 1/4h.- 3/16)e. 2h. ,
a general solution - w. ( h.) \u003d S. 1 e. 3h. + S. 2 e. - 2h. + (- 1/4h.- 3/16)e. 2h. .
Note 2.In the case when the Cauchy problem is set for an inhomogeneous equation, you must first find the general solution of the equation
w.( h.) = ,
deciding all the numeric values \u200b\u200bof the coefficients in w.*(h.). Then take advantage of the initial conditions and, substituting them in a general solution (and not in y *(h.)), find the values \u200b\u200bof the constant C I..
PRI ME R 13. Find the solution of the Cauchy problem:
w.¢ ¢ (h.) - W.¢ (h.) - 6w.(h.) = xe. 2x. , U.(0) = 0, U. ¢ (h.) = 0.
Decision. General solution of this equation
w.(h.) \u003d S. 1 e. 3h. + S. 2 e. - 2h. + (- 1/4h.- 3/16)e. 2h.
it was found in Example 12. To find a private solution that satisfies the initial conditions for the Cauchy problem, we obtain a system of equations
Deciding it we have C. 1 = 1/8, C. 2 \u003d 1/16. Consequently, the solution of the Cauchy problem is a function
w.(h.) = 1/8e. 3h. + 1/16e. - 2h. + (- 1/4h.- 3/16)e. 2h. .
Note 3.(superposition principle). If in the linear equation L N.[y.(x.)] \u003d F.(x.), where f.(x.) \u003d F. 1 (x.) + F. 2 (x.) I. y * 1 (x.) - solution equation L N.[y.(x.)] \u003d F. 1 (x.), but y * 2 (x.) - solution equation L N.[y.(x.)] \u003d F. 2 (x.), that function y *(h.) \u003d y * 1 (x.) + * 2 (x.) is an by solving the equation L N.[y.(x.)] \u003d F.(x.).
PRI MERE P14. Specify the type of general solution of the linear equation
w.¢ ¢ (h.) + 4w.(h.) \u003d x + sinx.
Decision. General solution of the corresponding homogeneous equation
` w.(x.) \u003d S. 1 cos.2x + S. 2 sin.2x.,
as a characteristic equation k. 2 + 4 = 0 has roots k. 1, 2 = ± 2i.. After a part of the equation does not correspond to the formula (17), but if you enter the notation f. 1 (x.) \u003d H., f. 2 (x.) \u003d SINX. and take advantage of the superposition principle , That particular solution of the inhomogeneous equation can be found as y *(h.) \u003d y * 1 (x.) + * 2 (x.), where y * 1 (x.) - solution equation W.¢ ¢ (h.) + 4w.(h.) \u003d H., but y * 2 (x.) - solution equation w.¢ ¢ (h.) + 4w.(h.) \u003d sinx. By formula (18)
y * 1 (x.) \u003d Ah + in,y * 2 (x.) \u003d Ssosx + dsinx.
Then a particular solution
y *(h.) \u003d Ah + in + ssosx + dsinx,
consequently, the general solution has the form
w.(h.) \u003d S. 1 cos.2x + S. 2 e. - 2h. + A. x + B + ssosx + dsinx.
PRI ME R15. The electrical circuit consists of a sequentially connected current source with EDC e.(t.) \u003d E SINw. t, inductance L.and tank FROM , and
Establishment of Education "Belarusian State
agricultural Academy"
Department of Higher Mathematics
Methodical instructions
according to the topics of the "linear differential equations of second order" by students of the accounting faculty of correspondence form of education (NEPO)
Gorki, 2013.
Linear differential equations
second order with constantcoefficients
Linear homogeneous differential equations
Linear second-order differential equation with constant coefficients Called the view equation
those. The equation that contains the desired function and its derivatives only in the first degree and does not contain their works. In this equation 
and 
- Some numbers, and the function 
set at some interval 
.
If a 
at the interval 
, then equation (1) will take a view
,
(2)
and called linear homogeneous . Otherwise, equation (1) is called linear inhomogeneous .
Consider a comprehensive function
,
(3)
where 
and 
- Valid functions. If the function (3) is a comprehensive solution of equation (2), then the actual part 
and imaginary part 
solutions 
separately are the solutions of the same homogeneous equation. Thus, any comprehensive solution of equation (2) generates two valid solutions to this equation.
Solutions of a homogeneous linear equation have properties:
If a 
there is a solution of equation (2), then the function 
where FROM - arbitrary constant will also be a solution of equation (2);
If a 
and 
there are solutions of equation (2), then the function 
will also be a solution equation (2);
If a 
and 
there are solutions of equation (2), then their linear combination 
will also be a solution equation (2), where 
and 
- Arbitrary constant.
Functions 
and 
called linearly dependent
At the interval 
if there are such numbers 
and 
, not equal zero At the same time, that equality is performed on this interval.
If equality (4) takes place only when 
and 
, then functions 
and 
called linearly independent
At the interval 
.
Example 1.
. Functions 
and 
linearly dependent because 
on the whole numeric straight. In this example 
.
Example 2.
. Functions 
and 
linearly independent at any interval, because equality 
possible only in the case when 
, I. 
.
Building a general solution of linear homogeneous
equations
In order to find the general solution of equation (2), you need to find two of its linearly independent decisions. 
and 
. Linear combination of these solutions 
where 
and 
- arbitrary constant, and will give a general solution of a linear homogeneous equation.
Linearly independent solutions of equation (2) will be signed as
,
(5)
where 
- Some number. Then 
,
. Substitute these expressions in equation (2):
or 
.
As 
T. 
. Thus, the function 
will be the solution of equation (2) if 
will satisfy the equation
.
(6)
Equation (6) is called characteristic equation For equation (2). This equation is an algebraic square equation.
Let be 
and 
there are roots of this equation. They can be either valid and different, or complex, or valid and equal. Consider these cases.
Let roots 
and 
the characteristic equation is valid and different. Then solutions of equation (2) will be functions 
and 
. These solutions are linearly independent, since equality 
can be performed only when 
, I. 
. Therefore, the general solution of equation (2) has the form
,
where 
and 
- Arbitrary constant.
Example 3.
.
Decision
. The characteristic equation for this differential will be 
. Solving it quadratic equation, find his roots 
and 
. Functions 
and 
are solutions of a differential equation. The general solution of this equation has the form 
.
Integrated number 
called the expression of the view 
where 
and 
- Actual numbers, and 
called an imaginary unit. If a 
, then 
called purely imaginary. If 
, then 
identified with a valid number 
.
Number 
called a valid part of the integrated number, and 
- imaginary part. If two complex numbers differ from each other only by the image of the imaginary part, they are closed conjugate: 
,
.
Example 4.
. Solve square equation 
.
Decision
. Discriminant equation 
. Then. Similarly, 
. Thus, this square equation has conjugate complex roots.
Let the roots of the characteristic equation are complex, i.e. 
,
where 
. Solutions equation (2) can be written as 
,
or 
,
. According to Euler formulas
,
.
Then. As is known if the complex function is a solution of a linear homogeneous equation, the solutions of this equation are valid and the imaginary parts of this function. Thus, the solutions of equation (2) will be functions 
and 
. Since equality
can only be executed if 
and 
, these solutions are linearly independent. Consequently, the general solution of equation (2) has the form
where 
and 
- Arbitrary constant.
Example 5.
. Find a general solution of differential equation 
.
Decision
. The equation 
it is characteristic for this differential. I solve it and get complex roots 
,
. Functions 
and 
are linearly independent solutions of a differential equation. The general solution of this equation has the form.
Let the roots of the characteristic equation are valid and equal, i.e. 
. Then the solutions of equation (2) are functions 
and 
. These solutions are linearly independent, since expressions be identically equal to zero only when 
and 
. Consequently, the general solution of equation (2) has the form 
.
Example 6.
. Find a general solution of differential equation 
.
Decision
. Characteristic equation 
it has equal roots 
. In this case, linearly independent solutions of the differential equation are functions. 
and 
. The general solution has the form 
.
Inhomogeneous linear second-order differential equations with constant coefficients
and special the right part
The total solution of the linear inhomogeneous equation (1) is equal to the sum of the overall solution 
the corresponding homogeneous equation and any particular solution 
inhomogeneous equation: 
.
In some cases, the private solution of the inhomogeneous equation can be found quite just in appearance of the right 
equations (1). Consider cases when possible.
those. The right side of the inhomogeneous equation is a polynomial degree m.. If a 
it is not the root of the characteristic equation, the particular solution of the inhomogeneous equation should be seen as a polynomial m..
Factors 
defined in the process of finding a private solution.
If 
it is the root of the characteristic equation, the private solution of the inhomogeneous equation should be sought in the form of
Example 7.
. Find a general solution of differential equation 
.
Decision
. Appropriate homogeneous equation for of this equation is an 
. His characteristic equation 
has roots 
and 
. The general solution of the homogeneous equation has the form 
.
As 
it is not a root of a characteristic equation, the particular solution of the inhomogeneous equation will be signed as a function 
. Find derivatives of this function 
,
and we substitute them in this equation:
or . We equate the coefficients for 
and free members: 
Solving this system, 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 overall solution of the corresponding homogeneous equation and a particular solution of heterogeneous: 
.
Let the heterogeneous equation relates
If a 
it is not a root of the characteristic equation, the particular solution of the inhomogeneous equation should be signed in the form. If 
there is a root of the characteristic equation of multiplicity k.
(k.\u003d 1 or k.\u003d 2), in this case, the particular solution of the inhomogeneous equation will be viewed.
Example 8.
. Find a general solution of differential equation 
.
Decision
. The characteristic equation for the corresponding homogeneous equation has the form 
. His roots 
,
. In this case, the general solution of the corresponding homogeneous equation is recorded as 
.
Since the number 3 is not the root of the characteristic equation, the particular solution of the inhomogeneous equation should be sought in the form of 
. Find the derivatives of the first and second orders :,
Substitute to the differential equation: 
+
+,
+,.
We equate the coefficients for 
and free members:
From here 
,
. Then the particular solution of this equation has the form 
, and the general decision
.
Lagrange Method Variations Arbitrary Permanent
The method of variation of arbitrary constants can be applied to any non-uniform linear equation with constant coefficients, regardless of the type of the right part. This method allows you to always find a general solution of an inhomogeneous equation, if a general solution is known to the corresponding homogeneous equation.
Let be 
and 
are linearly independent solutions of equation (2). Then the general solution of this equation is 
where 
and 
- Arbitrary constant. The essence of the method of variation of arbitrary constants is that the general solution of the equation (1) is searched as
where 
and 
- New unknown features that need to be found. Since unknown functions are two, then there are two equations containing these functions for their findings. These two equations make up the system
which is a linear algebraic system of equations relative to 
and 
. Solving this system, we will find 
and 
. Integrating both parts of the equalities received, find
and 
.
Substituting these expressions in (9), we obtain a general solution of a non-uniform linear equation (1).
Example 9.
. Find a general solution of differential equation 
.
Decision.
The characteristic equation for a homogeneous equation corresponding to this differential equation is 
. The roots of its complex 
,
. As 
and 
T. 
,
, and the general solution of a homogeneous equation has the form. Then the general solution of this inhomogeneous equation will be sought in the form where 
and 
- Unknown functions.
The system of equations for finding these unknown functions is
Deciding this system, we will find 
,
. Then
,
. Substitute the received expressions in the general decision formula:
This is the general solution of this differential equation obtained by the Lagrange method.
Questions for self-controlling knowledge
What differential equation is called a second-order linear differential equation with constant coefficients?
What a linear differential equation is called homogeneous, and what is heterogeneous?
What properties is the linear homogeneous equation?
What equation is called characteristic for a linear differential equation and how is it obtained?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients in the case of different roots of the characteristic equation?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients in the case of equal roots of the characteristic equation?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients in the case of complex roots of the characteristic equation?
How is the general solution of a linear inhomogeneous equation written?
In what form is a private solution of a linear inhomogeneous equation, if the roots of the characteristic equation are different and are not equal to zero, and the right side of the equation is a polynomial degree m.?
In what form is a private solution of a linear inhomogeneous equation, if among the roots of the characteristic equation there is one zero, and the right part of the equation is a polynomial m.?
What is the essence of the Lagrange method?
This article discloses the question of solving linear inhomogeneous second-order differential equations with constant coefficients. The theory will be considered together with the examples of the given tasks. For decryption incomprehensible terms It is necessary to contact the topic of the basic definitions and concepts of the theory of differential equations.
Consider a linear differential equation (LFD) of the second order with constant coefficients of the form y "" + p · y "+ q · y \u003d f (x), where the arbitrary numbers are p and q, and the existing function f (x) is continuous on the integration interval x.
Let us turn to the wording of the LFD general decision theorem.
Yandex.rtb R-A-339285-1
General decision theorem LDNU
Theorem 1.A general solution that is at the interval of the inhomogeneous differential equation of the form y (n) + F n - 1 (x) · y (n - 1) +. . . + F 0 (x) · y \u003d F (x) with continuous integration coefficients on the x interval F 0 (x), F 1 (x) ,. . . , f n - 1 (x) and continuous function F (x) is equal to the sum of the overall solution y 0, which corresponds to a log and some particular solution y ~, where the initial inhomogeneous equation is y \u003d y 0 + y ~.
It can be seen that the solution of such an equation of the second order has the form y \u003d y 0 + y ~. The algorithm of finding Y 0 is considered in the article on linear homogeneous second-order differential equations with constant coefficients. After that, you should move to the definition Y ~.
The choice of the LFD private solution depends on the view of the existing function f (x), located in the right part of the equation. To do this, it is necessary to consider separately solutions of linear inhomogeneous second-order differential equations with constant coefficients.
When f (x) is considered for a polynomial N-degree f (x) \u003d p n (x), it follows that the LFD's private solution is found according to the formula y ~ \u003d Q n (x) · x γ, where Q n ( x) is a polynomial degree n, R is the number of zero roots of the characteristic equation. The value y ~ is a private solution Y ~ "" + p · y ~ "+ q · y ~ \u003d f (x), then the available coefficients that are defined by the polynomial
Q n (x), find it with the help of the method of uncertain coefficients from the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 1.
Calculate on the Cauchy theorem y "" - 2 y "\u003d x 2 + 1, y (0) \u003d 2, y" (0) \u003d 1 4.
Decision
In other words, it is necessary to move to a private solution of the linear inhomogeneous differential equation of the second order with constant coefficients Y "" - 2 y "\u003d x 2 + 1, which will satisfy the specified conditions y (0) \u003d 2, y" (0) \u003d 1 4 .
The general solution of the linear inhomogeneous equation is the sum of the overall solution, which corresponds to the equation y 0 or the private solution of the inhomogeneous Y ~ equation, that is, y \u003d y 0 + y ~.
To begin with, we will find a general solution for the LFD, and after which is private.
Let us turn to finding y 0. Recording a characteristic equation will help to find the roots. We get that
k 2 - 2 k \u003d 0 k (k - 2) \u003d 0 k 1 \u003d 0, k 2 \u003d 2
Received that the roots are different and valid. So write
y 0 \u003d C 1 E 0 X + C 2 E 2 X \u003d C 1 + C 2 E 2 x.
We find y ~. It can be seen that the right side specified equation It is a polynomial of a second degree, then one of the roots is zero. From here we get that a private solution for Y ~ will
y ~ \u003d q 2 (x) · x γ \u003d (a x 2 + b x + c) · x \u003d a x 3 + b x 2 + c x, where the values \u200b\u200bA, B, C are taken by indefinite coefficients.
We find them from the equality of the form y ~ "" - 2 y ~ "\u003d x 2 + 1.
Then we get that:
y ~ "" - 2 y ~ "\u003d x 2 + 1 (a x 3 + b x 2 + c x)" "- 2 (a x 3 + b x 2 + c x)" \u003d x 2 + 1 3 a x 2 + 2 b x + c "- 6 a x 2 - 4 b x - 2 c \u003d x 2 + 1 6 a x + 2 b - 6 a x 2 - 4 b x - 2 c \u003d x 2 + 1 - 6 A x 2 + x (6 A - 4 B) + 2 B - 2 C \u003d x 2 + 1
Equating the coefficients with the same indicators of degrees X, we obtain a system of linear expressions - 6 a \u003d 1 6 A - 4 B \u003d 0 2 B - 2 C \u003d 1. When solving any of the ways, we find the coefficients and write: a \u003d - 1 6, b \u003d - 1 4, c \u003d - 3 4 and y ~ \u003d a x 3 + b x 2 + c x \u003d - 1 6 x 3 - 1 4 x 2 - 3 4 x.
This entry is called the general solution of the original linear inhomogeneous differential equation of the second order with constant coefficients.
To find a private solution that satisfies the conditions Y (0) \u003d 2, Y "(0) \u003d 1 4, it is required to determine the values C 1 and C 2.Based on the equality of the form y \u003d C 1 + C 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x.
We get that:
y (0) \u003d C 1 + C 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 xx \u003d 0 \u003d C 1 + C 2 y "(0) \u003d C 1 + C 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x "x \u003d 0 \u003d 2 c 2 e 2 x - 1 2 x 2 + 1 2 x + 3 4 x \u003d 0 \u003d 2 C 2 - 3 4
We work with the obtained system of equations of the form C 1 + C 2 \u003d 2 2 C 2 - 3 4 \u003d 1 4, where C 1 \u003d 3 2, C 2 \u003d 1 2.
Applying Cauchy theorem, we have that
y \u003d C 1 + C 2 E 2 X - 1 6 x 3 + 1 4 x 2 + 3 4 x \u003d 3 2 + 1 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x
Answer: 3 2 + 1 2 E 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x.
When the function f (x) is presented in the form of a piece of polynomial with a degree n and exponent F (x) \u003d p n (x) · EAX, then it turns out that the equation of the form y ~ \u003d EAX · Q n ( x) · X γ, where Q n (x) is a polynomial of n-degree, and R is the number of roots of the characteristic equation equal to α.
The coefficients belonging to Q n (x) are on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 2.
Find the general solution of the differential equation of the form y "" - 2 y "\u003d (x 2 + 1) · e x.
Decision
The equation general view y \u003d y 0 + y ~. Specified equation corresponds to the log y "" - 2 y "\u003d 0. According to the previous example, it can be seen that its roots are equal K 1 \u003d 0 and k 2 \u003d 2 and y 0 \u003d C 1 + C 2 E 2 x on a characteristic equation.
It can be seen that the right part of the equation is x 2 + 1 · e x. From here, the LFD is located through Y ~ \u003d E A x · Q n (x) · x γ, where q n (x), which is a polynomial of a second degree, where α \u003d 1 and r \u003d 0, because there is no root equal to 1. From here we get that
y ~ \u003d E A x · Q n (x) · x γ \u003d e x · a x 2 + b x + c · x 0 \u003d e x · a x 2 + b x + c.
A, B, C are unknown coefficients that can be found on the equality y ~ "" - 2 y ~ "\u003d (x 2 + 1) · e x.
Received that
y ~ "\u003d ex · a x 2 + b x + c" \u003d ex · a x 2 + b x + c + ex · 2 a x + b \u003d ex · a x 2 + x 2 a + b + b + C y ~ "" \u003d ex · a x 2 + x 2 a + b + b + c "\u003d \u003d ex · a x 2 + x 2 a + b + b + c + ex · 2 a x + 2 a + b \u003d \u003d EX · A x 2 + x 4 A + B + 2 A + 2 B + C
y ~ "" - 2 y ~ "\u003d (x 2 + 1) · ex ⇔ ex · a x 2 + x 4 a + b + 2 a + 2 b + c - - 2 ex · a x 2 + x 2 a + B + b + c \u003d x 2 + 1 · ex ⇔ ex · a x 2 - b x + 2 a - c \u003d (x 2 + 1) · ex ⇔ - a x 2 - b x + 2 A - C \u003d x 2 + 1 ⇔ - a x 2 - b x + 2 a - c \u003d 1 · x 2 + 0 · x + 1
Indicators with the same coefficients equating and get the system linear equations. From here and find A, B, C:
A \u003d 1 - b \u003d 0 2 A - C \u003d 1 ⇔ A \u003d - 1 B \u003d 0 C \u003d - 3
Answer: It can be seen that y ~ \u003d ex · (a x 2 + b x + c) \u003d ex · - x 2 + 0 · x - 3 \u003d - ex · x 2 + 3 is a special solution of the LFD, and y \u003d y 0 + y \u003d C 1 E 2 x - Ex И x 2 + 3 is a general solution for inhomogeneous second-order dyferal.
When the function is written as f (x) \u003d a 1 cos (β x) + b 1 sin β x, and A 1. and IN 1are numbers, then the LDDa equation y ~ \u003d A cos β x + b sin β x · x γ is considered to be a private solution of the LLD, where A and B are considered uncertain coefficients, and R by the number of complex conjugate roots belonging to a characteristic equation equal to ± I β . In this case, the search for coefficients is carried out on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 3.
Find a general solution of the differential equation of the form y "" + 4 y \u003d cos (2 x) + 3 sin (2 x).
Decision
Before writing a characteristic equation, we find Y 0. Then
k 2 + 4 \u003d 0 k 2 \u003d - 4 k 1 \u003d 2 i, k 2 \u003d - 2 i
We have a couple of comprehensively conjugate roots. We transform and get:
y 0 \u003d E 0 · (C 1 COS (2 x) + C 2 Sin (2 x)) \u003d C 1 COS 2 X + C 2 SIN (2 x)
The roots from the characteristic equation are considered to be a conjugate pair ± 2 i, then f (x) \u003d cos (2 x) + 3 sin (2 x). It can be seen that the search y ~ will be made from Y ~ \u003d (a COS (β x) + b sin (β x) · x γ \u003d (A COS (2 x) + b sin (2 x)) · x. Unknown The coefficients A and B will be seen from the equality of the form y ~ "" + 4 y ~ \u003d cos (2 x) + 3 sin (2 x).
We transform:
y ~ "\u003d ((a COS (2 x) + b sin (2 x) · x)" \u003d \u003d (- 2 A sin (2 x) + 2 b cos (2 x)) · X + A COS (2 x) + b sin (2 x) y ~ "" \u003d ((- 2 a sin (2 x) + 2 b cos (2 x)) · x + a COS (2 x) + b sin (2 x)) "\u003d \u003d (- 4 A COS (2 x) - 4 b sin (2 x)) · x - 2 A sin (2 x) + 2 b cos (2 x) - - 2 a sin (2 x) + 2 B cos (2 x) \u003d \u003d (- 4 A cos (2 x) - 4 b sin (2 x)) · x - 4 a sin (2 x) + 4 b cos (2 x)
Then it can be seen that
y ~ "" + 4 y ~ \u003d cos (2 x) + 3 sin (2 x) ⇔ (- 4 A COS (2 x) - 4 b sin (2 x)) · x - 4 a sin (2 x) + 4 B cos (2 x) + + 4 (A COS (2 x) + b sin (2 x)) · x \u003d cos (2 x) + 3 sin (2 x) ⇔ - 4 A SIN (2 x) + 4 B cos (2 x) \u003d cos (2 x) + 3 sin (2 x)
It is necessary to equate the coefficients of sinuses and cosine. We get a system of type:
4 a \u003d 3 4 b \u003d 1 ⇔ a \u003d - 3 4 b \u003d 1 4
It follows that y ~ \u003d (a COS (2 x) + b sin (2 x) · x \u003d - 3 4 cos (2 x) + 1 4 sin (2 x) · x.
Answer:the general solution of the initial LFD of the second order with constant coefficients is considered
y \u003d Y 0 + Y ~ \u003d C 1 COS (2 x) + C 2 SIN (2 x) + - 3 4 COS (2 x) + 1 4 sin (2 x) · x
When f (x) \u003d eax · p n (x) sin (β x) + q k (x) cos (β x), then y ~ \u003d eax · (L m (x) sin (β x) + n m (x) cos (β x) · x γ. We have that R is the number of comprehensively conjugate pairs of roots relating to the characteristic equation equal to α ± i β, where p n (x), q k (x), L m (x) and N M (x)are polynomials of degree n, k, t, m, where M \u003d M a x (n, k). Finding coefficients L M (X) and N M (x) It is performed based on the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 4.
Find a general solution y "" + 3 y "+ 2 y \u003d - E 3 x · ((38 x + 45) sin (5 x) + (8 x - 5) COS (5 x)).
Decision
Under the condition it can be seen that
α \u003d 3, β \u003d 5, p n (x) \u003d - 38 x - 45, q k (x) \u003d - 8 x + 5, n \u003d 1, k \u003d 1
Then m \u003d m a x (n, k) \u003d 1. We produce the finding Y 0, pre-writing the characteristic equation of the form:
k 2 - 3 k + 2 \u003d 0 d \u003d 3 2 - 4 · 1 · 2 \u003d 1 k 1 \u003d 3 - 1 2 \u003d 1, k 2 \u003d 3 + 1 2 \u003d 2
Received that the roots are valid and different. Hence y 0 \u003d C 1 E x + C 2 E 2 x. Next, it is necessary to look for a general decision, based on the inhomogeneous equation y ~ type
y ~ \u003d e α x · (L m (x) sin (β x) + n m (x) cos (β x) · x γ \u003d e 3 x · ((a x + b) COS (5 x) + (C x + d) sin (5 x)) · x 0 \u003d \u003d e 3 x · ((a x + b) cos (5 x) + (C x + d) sin (5 x))
It is known that A, B, C are coefficients, R \u003d 0, because there is no pair of conjugate roots belonging to the characteristic equation with α ± i β \u003d 3 ± 5 · i. These coefficients find from the equality obtained:
y ~ "" - 3 y ~ "+ 2 y ~ \u003d - E 3 x ((38 x + 45) sin (5 x) + (8 x - 5) cos (5 x)) ⇔ (E 3 x (( A x + b) COS (5 x) + (C x + d) sin (5 x))) "" - - 3 (E 3 x ((a x + b) COS (5 x) + (C x + D) sin (5 x))) \u003d - E 3 x ((38 x + 45) sin (5 x) + (8 x - 5) COS (5 x))
Finding derivative and similar components gives
E 3 x · ((15 A + 23 C) · X · Sin (5 x) + + (10 A + 15 B - 3 C + 23 D) · Sin (5 x) + + (23 A - 15 C) · X · COS (5 x) + (- 3 A + 23 B - 10 C - 15 D) · COS (5 x)) \u003d \u003d - E 3 x · (38 · x · sin (5 x) + 45 · SIN (5 x) + + 8 · x · COS (5 x) - 5 · COS (5 x))
After equating coefficients, we get a system of type
15 A + 23 C \u003d 38 10 A + 15 B - 3 C + 23 D \u003d 45 23 A - 15 C \u003d 8 - 3 A + 23 B - 10 C - 15 D \u003d - 5 ⇔ A \u003d 1 B \u003d 1 C \u003d 1 d \u003d 1
It follows from everything that
y ~ \u003d E 3 x · ((a x + b) cos (5 x) + (C x + d) sin (5 x)) \u003d \u003d E 3 x · ((x + 1) COS (5 x) + (x + 1) sin (5 x))
Answer:now the total solution of the specified linear equation is obtained:
y \u003d y 0 + y ~ \u003d \u003d c 1 e x + c 2 e 2 x + e 3 x · ((x + 1) cos (5 x) + (x + 1) sin (5 x))
Algorithm Decision LDNU
Definition 1.Any other type of function f (x) to solve compliance with the solution algorithm:
- finding a general solution of the corresponding linear homogeneous equation, where Y 0 \u003d C 1 ⋅ Y 1 + C 2 ⋅ Y 2, where Y 1. and Y 2.are linearly independent private solutions With 1. and With 2are considered arbitrary constant;
- the adoption as a general solution of the LFD y \u003d C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2;
- determination of derivatives through a system of type C 1 "(x) + y 1 (x) + C 2" (x) · y 2 (x) \u003d 0 C 1 "(x) + y 1" (x) + C 2 " (x) · y 2 "(x) \u003d f (x), and finding functions C 1 (X) and C 2 (x) by integration.
Example 5.
Find a general solution for Y "" + 36 y \u003d 24 sin (6 x) - 12 COS (6 x) + 36 E 6 x.
Decision
We turn to the writing of the characteristic equation, pre-writing Y 0, Y "" + 36 y \u003d 0. We write and resolve:
k 2 + 36 \u003d 0 k 1 \u003d 6 i, k 2 \u003d - 6 i ⇒ y 0 \u003d C 1 cos (6 x) + c 2 sin (6 x) ⇒ y 1 (x) \u003d cos (6 x), Y 2 (x) \u003d sin (6 x)
We have that the recording of a general solution of a given equation will be viewed y \u003d C 1 (x) · COS (6 x) + C 2 (x) · sin (6 x). It is necessary to go to the definition of derived functions. C 1 (X) and C 2 (x) By system with equations:
C 1 "(x) · COS (6 x) + C 2" (x) · sin (6 x) \u003d 0 C 1 "(x) · (COS (6 x))" + C 2 "(x) · (SIN (6 x)) "\u003d 0 ⇔ C 1" (X) · COS (6 x) + C 2 "(x) · sin (6 x) \u003d 0 C 1" (x) (- 6 sin (6 x) + C 2 "(x) (6 cos (6 x)) \u003d 24 sin (6 x) - 12 cos (6 x) + 36 E 6 x
It is necessary to make a decision regarding C 1 "(x) and C 2 "(x) Using any way. Then write:
C 1 "(x) \u003d - 4 sin 2 (6 x) + 2 sin (6 x) COS (6 x) - 6 E 6 x Sin (6 x) C 2" (x) \u003d 4 sin (6 x) COS (6 x) - 2 COS 2 (6 x) + 6 E 6 X COS (6 x)
Each equations should be integrated. Then we write the resulting equations:
C 1 (x) \u003d 1 3 sin (6 x) COS (6 x) - 2 x - 1 6 COS 2 (6 x) + + 1 2 E 6 x COS (6 x) - 1 2 E 6 X SIN ( 6 x) + C 3 C 2 (x) \u003d - 1 6 sin (6 x) cos (6 x) - x - 1 3 cos 2 (6 x) + + 1 2 E 6 x COS (6 x) + 1 2 E 6 X SIN (6 x) + C 4
It follows that the general decision will look at:
y \u003d 1 3 sin (6 x) COS (6 x) - 2 x - 1 6 COS 2 (6 x) + + 1 2 E 6 X COS (6 x) - 1 2 E 6 x Sin (6 x) + C 3 · COS (6 x) + + - 1 6 sin (6 x) COS (6 x) - x - 1 3 COS 2 (6 x) + + 1 2 E 6 X COS (6 x) + 1 2 E 6 x sin (6 x) + C 4 · sin (6 x) \u003d \u003d - 2 x · cos (6 x) - x · sin (6 x) - 1 6 COS (6 x) + + 1 2 E 6 x + C 3 · COS (6 x) + C 4 · SIN (6 x)
Answer: y \u003d y 0 + y ~ \u003d - 2 x · cos (6 x) - x · sin (6 x) - 1 6 cos (6 x) + + 1 2 E 6 x + C 3 · COS (6 x) + C 4 · SIN (6 x)
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