Differential equations with the right-hand side of a special form. Second order inhomogeneous differential equations
The lecture studies LNDE - linear inhomogeneous differential equations. The structure of the general solution is considered, the solution to the LNDE by the method of variation of arbitrary constants, the solution to the LNDE with constant coefficients and a special right-hand side. The issues under consideration are used in the study of forced oscillations in physics, electrical engineering and electronics, the theory of automatic control.
1. The structure of the general solution of a linear inhomogeneous differential equation of the second order.
Consider first a linear inhomogeneous equation of arbitrary order:
Taking into account the designation, you can write:
In this case, we will assume that the coefficients and the right-hand side of this equation are continuous on a certain interval.
Theorem. The general solution of a linear inhomogeneous differential equation in some domain is the sum of any of its solutions and the general solution of the corresponding linear homogeneous differential equation.
Evidence. Let Y be some solution to an inhomogeneous equation.
Then, substituting this solution into the original equation, we obtain the identity:
Let be 
- fundamental system of solutions of a linear homogeneous equation 
... Then the general solution of the homogeneous equation can be written in the form:
In particular, for a linear inhomogeneous differential equation of order 2, the structure of the general solution has the form:
where 
is the fundamental system of solutions of the corresponding homogeneous equation, and 
- any particular solution of an inhomogeneous equation.
Thus, to solve a linear inhomogeneous differential equation, it is necessary to find the general solution of the corresponding homogeneous equation and somehow find one particular solution of the inhomogeneous equation. It is usually found by selection. We will consider the methods of selecting a private solution in the following questions.
2. Method of variation
In practice, it is convenient to apply the method of variation of arbitrary constants.
To do this, first find the general solution of the corresponding homogeneous equation in the form:
Then, setting the coefficients C i functions from x, a solution to the inhomogeneous equation is sought:
One can prove that to find the functions C i (x) it is necessary to solve the system of equations:
Example. Solve the equation 
Solving the linear homogeneous equation 
The solution to the inhomogeneous equation will have the form:
We compose a system of equations:
Let's solve this system:
From the relation we find the function Oh).
Now we find B (x).
We substitute the obtained values \u200b\u200binto the formula for the general solution of the inhomogeneous equation:
Final answer: 
Generally speaking, the method of variation of arbitrary constants is suitable for finding solutions to any linear inhomogeneous equation. But since Finding the fundamental system of solutions to the corresponding homogeneous equation can be a rather difficult task, this method is mainly used for inhomogeneous equations with constant coefficients.
3. Equations with the right-hand side of a special form
It seems possible to represent the form of a particular solution depending on the form of the right-hand side of the inhomogeneous equation.
The following cases are distinguished:
I. The right side of the linear inhomogeneous differential equation has the form:
where is a polynomial of degree m.
Then a particular solution is sought in the form:
Here Q(x) is a polynomial of the same degree as P(x) , but with undefined coefficients, and r - a number showing how many times the number is the root of the characteristic equation for the corresponding linear homogeneous differential equation.
Example. Solve the equation 
.
Let's solve the corresponding homogeneous equation: 
Now we will find a particular solution of the original inhomogeneous equation.
Let us compare the right side of the equation with the view of the right side discussed above.
We are looking for a private solution in the form: 
where
Those. 
Now we define the unknown coefficients ANDand IN.
Let's substitute a particular solution in general form into the original inhomogeneous differential equation.
Total, a particular solution: 
Then the general solution of the linear inhomogeneous differential equation:
II. The right side of the linear inhomogeneous differential equation has the form:
Here R 1 (x)and R 2 (x) - polynomials of degree m 1 and m 2 respectively.
Then the particular solution of the inhomogeneous equation will have the form:
where the number r shows how many times the number 
is the root of the characteristic equation for the corresponding homogeneous equation, and Q 1
(x)
and
Q 2
(x)
- polynomials of degree at most mwhere m- the largest of the degrees m 1
and m 2
.
Summary table of types of private decisions
for different kinds of right hand sides
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The right side of the differential equation |
characteristic equation |
Types of private |
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1. The number is not a root of the characteristic equation |
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2. Number - the root of the characteristic equation of multiplicity |
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1. Number |
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2. Number |
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1. Numbers |
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2. Numbers |
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1. Numbers | |||
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2. Numbers |
is not a root of the characteristic equation
is the root of the characteristic equation of multiplicity 
are the roots of the characteristic equation of multiplicity 
are not roots of the characteristic equation of multiplicity 
are the roots of the characteristic equation of multiplicity 
Note that if the right-hand side of the equation is a combination of expressions of the form considered above, then the solution is found as a combination of solutions to auxiliary equations, each of which has a right-hand side corresponding to the expression included in the combination.
Those. if the equation has the form: 
, then a particular solution of this equation will be 
where at 1
and at 2
- particular solutions of auxiliary equations
and 
For illustration, let's solve the above example in a different way.
Example. Solve the equation 
We represent the right side of the differential equation as the sum of two functions f 1 (x) + f 2 (x) = x + (- sin x).
Let's compose and solve the characteristic equation:
We get: Ie 
Total: 
Those. the desired particular solution has the form: 
General solution to a non-homogeneous differential equation:
Let's consider examples of applying the described methods.
Example 1 .. Solve the equation 
Let us compose the characteristic equation for the corresponding linear homogeneous differential equation:
Now we will find a particular solution of the inhomogeneous equation in the form:
Let's use the method of undefined coefficients.
Substituting into the original equation, we get:
A particular solution looks like: 
General solution of a linear inhomogeneous equation: 
Example. Solve the equation 
Characteristic equation:
General solution to the homogeneous equation: 
A particular solution to an inhomogeneous equation: 
.
Find the derivatives and substitute them into the original inhomogeneous equation:
We obtain a general solution to the inhomogeneous differential equation:
Fundamentals of solving linear inhomogeneous differential equations of the second order (LNDU-2) with constant coefficients (PC)
The 2nd order LNDE with constant coefficients $ p $ and $ q $ has the form $ y "" + p \\ cdot y "+ q \\ cdot y \u003d f \\ left (x \\ right) $, where $ f \\ left (x \\ right) $ is a continuous function.
In relation to LNDU 2 with PC, the following two statements are true.
Suppose that some function $ U $ is an arbitrary particular solution of an inhomogeneous differential equation. Suppose also that some function $ Y $ is a general solution (GR) of the corresponding linear homogeneous differential equation (LDE) $ y "" + p \\ cdot y "+ q \\ cdot y \u003d 0 $. Then the GD of LDE-2 is equal to the sum of the indicated particular and general solutions, that is, $ y \u003d U + Y $.
If the right side of the 2nd order LNDE is a sum of functions, that is, $ f \\ left (x \\ right) \u003d f_ (1) \\ left (x \\ right) + f_ (2) \\ left (x \\ right) +. .. + f_ (r) \\ left (x \\ right) $, then first you can find the PD $ U_ (1), U_ (2), ..., U_ (r) $, which correspond to each of the functions $ f_ ( 1) \\ left (x \\ right), f_ (2) \\ left (x \\ right), ..., f_ (r) \\ left (x \\ right) $, and only after that write the LNDE-2 PD in the form $ U \u003d U_ (1) + U_ (2) + ... + U_ (r) $.
2nd order LNDU solution from PC
Obviously, the form of this or that PD $ U $ of a given LNDE-2 depends on the specific form of its right-hand side $ f \\ left (x \\ right) $. The simplest cases of searching for the PD LNDE-2 are formulated in the form of the following four rules.
Rule number 1.
The right side of the LNDU-2 has the form $ f \\ left (x \\ right) \u003d P_ (n) \\ left (x \\ right) $, where $ P_ (n) \\ left (x \\ right) \u003d 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 \u003d Q_ (n) \\ left (x \\ right) \\ cdot x ^ (r) $, where $ Q_ (n) \\ left (x \\ right) $ is another polynomial of that 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 zero. The coefficients of the polynomial $ Q_ (n) \\ left (x \\ right) $ are found by the method of undefined coefficients (NK).
Rule number 2.
The right side of the LNDU-2 has the form $ f \\ left (x \\ right) \u003d 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 \u003d Q_ (n) \\ left (x \\ right) \\ cdot x ^ (r) \\ cdot e ^ (\\ alpha \\ cdot x) $, where $ Q_ (n) \\ The coefficients of the polynomial $ Q_ (n) \\ left (x \\ right) $ are found by the NK method.
Rule number 3.
The right side of LNDU-2 is $ f \\ left (x \\ right) \u003d 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 \u003d \\ left (A \\ cdot \\ cos \\ left (\\ beta \\ cdot x \\ right) + B \\ cdot \\ sin \\ left (\\ beta \\ cdot x \\ right) \\ right ) \\ cdot x ^ (r) $, where $ A $ and $ B $ are unknown coefficients, and $ r $ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $ i \\ cdot \\ beta $. The coefficients $ A $ and $ B $ are found by the NK method.
Rule number 4.
The right side of the LNDE-2 is $ f \\ left (x \\ right) \u003d 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 \u003d 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 corresponding to LODE-2, equal to $ \\ alpha + i \\ cdot \\ beta $. The coefficients of the polynomials $ Q_ (s) \\ left (x \\ right) $ and $ R_ (s) \\ left (x \\ right) $ are found by the NK method.
The NDT method consists in applying the following rule. In order to find the unknown coefficients of the polynomial that are part of the particular solution of the inhomogeneous differential equation of the LNDE-2, it is necessary:
- substitute PD $ U $, written in general form, into the left side of LNDU-2;
- on the left side of LNDU-2, simplify and group members with the same powers of $ x $;
- in the resulting identity, equate the coefficients of the terms with the same powers $ x $ of the left and right sides;
- solve the resulting system of linear equations for unknown coefficients.
Example 1
Problem: find OR LNDU-2 $ y "" - 3 \\ cdot y "-18 \\ cdot y \u003d \\ left (36 \\ cdot x + 12 \\ right) \\ cdot e ^ (3 \\ cdot x) $. Find also PD satisfying the initial conditions $ y \u003d 6 $ for $ x \u003d 0 $ and $ y "\u003d 1 $ for $ x \u003d 0 $.
We write down the corresponding LODU-2: $ y "" - 3 \\ cdot y "-18 \\ cdot y \u003d 0 $.
Characteristic equation: $ k ^ (2) -3 \\ cdot k-18 \u003d 0 $. Roots of the characteristic equation: $ k_ (1) \u003d -3 $, $ k_ (2) \u003d 6 $. These roots are valid and distinct. Thus, the OR of the corresponding LODE-2 has the form: $ Y \u003d C_ (1) \\ cdot e ^ (- 3 \\ cdot x) + C_ (2) \\ cdot e ^ (6 \\ cdot x) $.
The right side of this LNDE-2 has the form $ \\ left (36 \\ cdot x + 12 \\ right) \\ cdot e ^ (3 \\ cdot x) $. It is necessary to consider the coefficient of the exponent of the exponent $ \\ alpha \u003d 3 $. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PD of this LNDE-2 has the form $ U \u003d \\ left (A \\ cdot x + B \\ right) \\ cdot e ^ (3 \\ cdot x) $.
We will search for the coefficients $ A $, $ B $ by the NK method.
We find the first derivative of the PD:
$ U "\u003d \\ left (A \\ cdot x + B \\ right) ^ ((")) \\ cdot e ^ (3 \\ cdot x) + \\ left (A \\ cdot x + B \\ right) \\ cdot \\ left ( e ^ (3 \\ cdot x) \\ right) ^ ((")) \u003d $
$ \u003d A \\ cdot e ^ (3 \\ cdot x) + \\ left (A \\ cdot x + B \\ right) \\ cdot 3 \\ cdot e ^ (3 \\ cdot x) \u003d \\ left (A + 3 \\ cdot A \\ We find the second derivative of the PD:
{!LANG-b6f71e3c37e26e7a639013e11567f773!}
$ U "" \u003d \\ 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) ^ ((")) \u003d $
$ \u003d 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) \u003d \\ left (6 \\ cdot A + 9 \\ cdot A \\ cdot x + 9 \\ cdot B \\ right) \\ cdot e ^ (3 \\ cdot x). $
Substitute functions $ U "" $, $ U "$ and $ U $ instead of $ y" "$, $ y" $ and $ y $ into the given LNDU-2 $ y "" - 3 \\ cdot y "-18 \\ cdot y \u003d \\ left (36 \\ cdot x + 12 \\ right) \\ cdot e ^ (3 \\ cdot x). $ In this case, since the exponent $ e ^ (3 \\ cdot x) $ enters as a factor in all components, then its can be omitted.
$ 6 \\ cdot A + 9 \\ cdot A \\ cdot x + 9 \\ cdot B-3 \\ cdot \\ left (A + 3 \\ cdot A \\ cdot x + 3 \\ cdot B \\ right) -18 \\ cdot \\ left (A \\ We perform the actions on the left side of the resulting equality:
$ -18 \\ cdot A \\ cdot x + 3 \\ cdot A-18 \\ cdot B \u003d 36 \\ cdot x + 12. $
We apply the NDT method. We get a system of linear equations with two unknowns:
$ -18 \\ cdot A \u003d 36; $
$ 3 \\ cdot A-18 \\ cdot B \u003d 12. $
The solution to this system is: $ A \u003d -2 $, $ B \u003d -1 $.
CR $ U \u003d \\ left (A \\ cdot x + B \\ right) \\ cdot e ^ (3 \\ cdot x) $ for our problem looks like this: $ U \u003d \\ left (-2 \\ cdot x-1 \\ right) \\ cdot e ^ (3 \\ cdot x) $.
OP $ y \u003d Y + U $ for our problem looks like this: $ y \u003d C_ (1) \\ cdot e ^ (- 3 \\ cdot x) + C_ (2) \\ cdot e ^ (6 \\ cdot x) + \\ To search for a PD satisfying the given initial conditions, we find the derivative $ y "$ OP:
$ y "\u003d - 3 \\ cdot C_ (1) \\ cdot e ^ (- 3 \\ cdot x) +6 \\ cdot C_ (2) \\ cdot e ^ (6 \\ cdot x) -2 \\ cdot e ^ (3 \\ Substitute in $ y $ and $ y "$ the initial conditions $ y \u003d 6 $ at $ x \u003d 0 $ and $ y" \u003d 1 $ at $ x \u003d 0 $:
$ 6 \u003d C_ (1) + C_ (2) -1; $
$ 1 \u003d -3 \\ cdot C_ (1) +6 \\ cdot C_ (2) -2-3 \u003d -3 \\ cdot C_ (1) +6 \\ cdot C_ (2) -5. $
We got a system of equations:
$ C_ (1) + C_ (2) \u003d 7; $
$ -3 \\ cdot C_ (1) +6 \\ cdot C_ (2) \u003d 6. $
We solve it. We find $ C_ (1) $ by Cramer's formula, and $ C_ (2) $ is determined from the first equation:
$ C_ (1) \u003d \\ frac (\\ left | \\ begin (array) (cc) (7) & (1) \\\\ (6) & (6) \\ end (array) \\ right |) (\\ left | \\ cdot 6- \\ left (-3 \\ right) \\ cdot 1) \u003d \\ frac (36) (9) \u003d 4; C_ (2) \u003d 7-C_ (1) \u003d 7-4 \u003d 3. $
Thus, the PD of this differential equation has the form: $ y \u003d 4 \\ cdot e ^ (- 3 \\ cdot x) +3 \\ cdot e ^ (6 \\ cdot x) + \\ left (-2 \\ cdot x-1 \\ right ) \\ cdot e ^ (3 \\ cdot x) $.
Second-order inhomogeneous differential equations with constant coefficients
General solution structure
A linear inhomogeneous equation of this type has the form:
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Where |
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p q {!LANG-b4105a8365f96fbf4ef90b1000a2102e!} {!LANG-9d7bf075372908f55e2d945c39e0a613!}, {!LANG-c3be117041a113540deb0ff532b19543!} - constant numbers (which can be both real and complex). For each such equation, the corresponding homogeneous equation:
Theorem: The general solution to the inhomogeneous equation is the sum of the general solution y 0 (x) of the corresponding homogeneous equation and particular solution y 1 (x) of the inhomogeneous equation:
Below we will consider two ways to solve nonhomogeneous differential equations. Constant variation method If the general solution y 0 of the associated homogeneous equation is known, then the general solution of the inhomogeneous equation can be found using method of variation of constants... Let the general solution of a homogeneous second order differential equation have the form: Instead of permanent C 1 and C 2 we will consider auxiliary functions C 1 (x) and C 2 (x). We will look for these functions so that the solution satisfies the inhomogeneous equation with the right-hand side f(x). Unknown functions C 1 (x) and C 2 (x) are determined from a system of two equations:
Undefined coefficient method Right part f(x) of a nonhomogeneous differential equation is often a polynomial, an exponential or trigonometric function, or some combination of these functions. In this case, it is more convenient to search for a solution using undefined coefficient method... We emphasize that this method works only for a limited class of functions on the right side, such as  In both cases, the choice of a particular solution must correspond to the structure of the right-hand side of the inhomogeneous differential equation. In case 1, if the number α in the exponential function coincides with the root of the characteristic equation, then the particular solution will contain an additional factor x s where s - multiplicity of the root α in the characteristic equation. In case 2, if the number α + βi coincides with the root of the characteristic equation, then the expression for the particular solution will contain an additional factor x... Unknown coefficients can be determined by substituting the found expression for a particular solution into the original inhomogeneous differential equation. Superposition principle If the right side of the inhomogeneous equation is the sum multiple functions like then a particular solution to the differential equation will also be the sum of particular solutions constructed separately for each term on the right-hand side. |
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Example 1 |
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Solve differential equation y "" + y \u003d sin (2 x). Decision. We first solve the corresponding homogeneous equation y "" + y \u003d 0. In this case, the roots of the characteristic equation are purely imaginary: Therefore, the general solution of the homogeneous equation is determined by the expression Let's go back to the inhomogeneous equation. We will seek its solution in the form using the method of variation of constants. Functions C 1 (x) and C 2 (x) can be found from the following system of equations:
Let us express the derivative C 1 " (x) from the first equation:
Substituting into the second equation, we find the derivative C 2 " (x):
Hence it follows that Integrating expressions for derivatives C 1 " (x) and C 2 " (x), we get: {!LANG-b4105a8365f96fbf4ef90b1000a2102e!} A 1 , A 2 - constants of integration. Now we substitute the found functions C 1 (x) and C 2 (x) into the formula for y 1 (x) and write down the general solution of the inhomogeneous equation:
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Example 2 |
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Find the general solution to the equation y "" + y " −6y = 36x. Decision. Let's use the method of undefined coefficients. The right side of the given equation is a linear function f(x) \u003d ax + b... Therefore, we will look for a particular solution in the form The derivatives are equal:
Substituting this into the differential equation, we get:
The last equation is an identity, that is, it is true for all x, therefore, we equate the coefficients of the terms with the same powers x on the left and right side: From the resulting system, we find: A = −6, B \u003d −1. As a result, the particular solution is written as Now we will find the general solution of the homogeneous differential equation. Let us calculate the roots of the auxiliary characteristic equation: Therefore, the general solution to the corresponding homogeneous equation has the form: So, the general solution of the original inhomogeneous equation is expressed by the formula |
General DE integral.
Solve differential equation
But the funny thing is that the answer is already known: more precisely, we must also add a constant: A general integral is a solution to a differential equation.
Method of variation of arbitrary constants. Solution examples
The method of variation of arbitrary constants is used to solve inhomogeneous differential equations. This lesson is intended for those students who are already more or less well versed in the topic. If you are just starting to get acquainted with DU, i.e. If you are a teapot, I recommend starting with the first lesson: First order differential equations. Solution examples... And if you are already finishing, please discard the possible preconceived notion that the method is difficult. Because it's simple.
In what cases is the method of variation of arbitrary constants applied?
1) The method of variation of an arbitrary constant can be used to solve linear non-uniform DE of the 1st order... Since the equation is of the first order, then the constant (constant) is also one.
2) The method of variation of arbitrary constants is used to solve some linear inhomogeneous equations of the second order... Two constants vary here.
It is logical to assume that the lesson will consist of two paragraphs…. I wrote this proposal, and for 10 minutes I was painfully thinking what other clever crap to add for a smooth transition to practical examples. But for some reason, there are no thoughts after the holidays, although he did not seem to abuse anything. Therefore, let's go straight to the first paragraph.
Variation method of an arbitrary constant for a linear inhomogeneous first order equation
Before considering the method of variation of an arbitrary constant, it is advisable to be familiar with the article Linear differential equations of the first order... In that lesson, we practiced first solution non-uniform DE of the 1st order. This first solution, I remind you, is called replacement method or bernoulli method (not to be confused with bernoulli equation!!!)
We will now consider second solution - method of variation of an arbitrary constant. I will give just three examples, and I will take them from the above lesson. Why so few? Because in fact the solution in the second way will be very similar to the solution in the first way. In addition, according to my observations, the method of variation of arbitrary constants is used less often than the method of replacement.
Example 1
Find the general solution to the differential equation (Diffur from Example # 2 of the lesson Linear inhomogeneous DE of the 1st order)
Decision: This equation is linear inhomogeneous and has a familiar form:
At the first stage, it is necessary to solve a simpler equation: That is, we stupidly zero the right side - instead of writing zero. The equation I will call auxiliary equation.
In this example, you need to solve the following auxiliary equation:
Before us separable equation, the solution of which (hopefully) is no longer difficult for you:
Thus: - general solution of the auxiliary equation.
In the second step replace a constant of some yet unknown function that depends on "x":
Hence the name of the method - we vary the constant. Alternatively, the constant can be some function that we have to find now.
IN original inhomogeneous equation, we will replace:
Substitute and into the equation:
Control moment - the two terms on the left cancel out... If this does not happen, you should look for the error above.
As a result of the replacement, an equation with separable variables is obtained. Separate variables and integrate.
What a blessing, exhibitors are also declining:
Add the "normal" constant to the found function:
At the final stage, we recall our replacement:
Function just found!
So the general solution is:
Answer: common decision: 
If you print out two solutions, you will easily notice that in both cases we found the same integrals. The only difference is in the solution algorithm.
Now for something more complicated, I will also comment on the second example:
Example 2
Find the general solution to the differential equation (Diffur from Example No. 8 of the lesson Linear inhomogeneous DE of the 1st order)
Decision: Let us bring the equation to the form:
Let us zero the right-hand side and solve the auxiliary equation:
We separate the variables and integrate: General solution of the auxiliary equation:
In the inhomogeneous equation, we replace:
According to the rule of product differentiation:
We also substitute in the original inhomogeneous equation:
The two terms on the left side are canceling, which means we are on the right track: 
We integrate by parts. The tasty letter from the formula for integration by parts has already been used in the solution, so we use, for example, the letters "a" and "be":
Eventually:
Now let's remember the replacement:
Answer: common decision:
Variation method of arbitrary constants for a linear inhomogeneous second order equation with constant coefficients
We often heard the opinion that the method of variation of arbitrary constants for a second-order equation is not an easy thing. But I guess the following: most likely, the method seems difficult to many, since it is not so common. But in reality, there are no particular difficulties - the course of the decision is clear, transparent, understandable. And beautiful.
To master the method, it is desirable to be able to solve inhomogeneous second-order equations by selecting a particular solution in the form of the right-hand side. This method is discussed in detail in the article. 2nd order inhomogeneous DE... We recall that the second-order linear inhomogeneous equation with constant coefficients has the form:
The selection method, which was considered in the above lesson, works only in a limited number of cases when polynomials, exponents, sines, cosines are on the right side. But what to do when on the right, for example, fraction, logarithm, tangent? In such a situation, the method of variation of constants comes to the rescue.
Example 4
Find the general solution of the second order differential equation 
Decision: There is a fraction on the right side of this equation, so we can immediately say that the method of selecting a particular solution does not work. We use the method of variation of arbitrary constants.
Nothing foreshadows a thunderstorm, the beginning of the solution is completely ordinary:
Find common decision corresponding homogeneous equations:
Let's compose and solve the characteristic equation: 
- conjugate complex roots are obtained, so the general solution is:
Pay attention to the record of the general solution - if there are brackets, then we expand them.
Now we do practically the same trick as for the first-order equation: we vary the constants, replacing them with unknown functions. I.e, general solution to heterogeneouswe will seek equations in the form:
Where - yet unknown functions.
It looks like a landfill for household waste, but now we will sort everything.
Derivatives of functions act as unknowns. Our goal is to find derivatives, and the found derivatives must satisfy both the first and second equations of the system.
Where do the "games" come from? The stork brings them. We look at the general solution obtained earlier and write down:
Find the derivatives:
With the left parts sorted out. What's on the right?
Is the right side of the original equation, in this case:
This article reveals the question of solving linear inhomogeneous differential equations of the second order with constant coefficients. The theory will be considered together with examples of the problems presented. To decipher the incomprehensible terms, it is necessary to refer to the topic of the basic definitions and concepts of the theory of differential equations.
Consider a second-order linear differential equation (LDDE) with constant coefficients of the form y "" + p y "+ q y \u003d f (x), where p and q are arbitrary numbers, and the existing function f (x) is continuous on the integration interval x.
Let us turn to the formulation of the theorem for the general solution of the LNDE.
Yandex.RTB R-A-339285-1
General solution theorem for LDNU
Theorem 1The 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 \u003d f (x) with continuous integration coefficients on the x interval f 0 (x), f 1 (x),. ... ... , f n - 1 (x) and the continuous function f (x) is equal to the sum of the general solution y 0, which corresponds to the LODE and some particular solution y ~, where the original inhomogeneous equation is y \u003d y 0 + y ~.
Hence, it is seen that the solution of such a second-order equation has the form y \u003d y 0 + y ~. The algorithm for finding y 0 is considered in the article on linear homogeneous differential equations of the second order with constant coefficients. Then one should proceed to the definition of y ~.
The choice of a particular solution to the LNDE depends on the form of the existing function f (x) located on the right side of the equation. To do this, it is necessary to consider separately the solutions of linear inhomogeneous differential equations of the second order with constant coefficients.
When f (x) is considered to be a polynomial of degree n f (x) \u003d P n (x), it follows that the particular solution of the LNDE is found by a formula of the form y ~ \u003d 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 ~ \u003d f (x), then the available coefficients, which are defined by the polynomial
Q n (x), we find using the method of undefined coefficients from the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 1
Calculate by Cauchy's 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 pass to a particular solution of a linear inhomogeneous differential equation of the second order with constant coefficients y "" - 2 y "\u003d x 2 + 1, which will satisfy the given conditions y (0) \u003d 2, y" (0) \u003d 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 \u003d y 0 + y ~.
First, let's find a general solution for the LNDE, and then a particular one.
Let's move on to finding y 0. Writing the characteristic equation will help you 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
Got the roots different and valid. Therefore, we write
y 0 \u003d C 1 e 0 x + C 2 e 2 x \u003d C 1 + C 2 e 2 x.
Find y ~. It is 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 the particular solution for y ~ will be
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\u200bof A, B, C take undefined coefficients.
Let us find them from an 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 exponents of 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 by any of the methods, we find the coefficients and write down: 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 record is called the general solution of the original linear inhomogeneous differential equation of the second order with constant coefficients.
To find a particular solution that satisfies the conditions y (0) \u003d 2, y "(0) \u003d 1 4, it is required to determine the values C 1 and C 2based on an equality of the form y \u003d C 1 + C 2 e 2 x - 1 6 x 3 + 1 4 x 2 + 3 4 x.
We get that:
y (0) \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 \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 resulting 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 the 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 \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 represented as a product of a polynomial with degree n and an exponent f (x) \u003d P n (x) eax, then from this we obtain that an equation of the form y ~ \u003d eax Q n ( x) x γ, where Q n (x) is a polynomial of degree n, and r is the number of roots of the characteristic equation equal to α.
The coefficients belonging to Q n (x) are found by the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 2
Find the general solution of a differential equation of the form y "" - 2 y "\u003d (x 2 + 1) · e x.
Decision
General equation y \u003d y 0 + y ~. The above equation corresponds to the LODE y "" - 2 y "\u003d 0. From the previous example, you can see that its roots are k 1 \u003d 0 and k 2 \u003d 2 and y 0 \u003d C 1 + C 2 e 2 x according to the characteristic equation.
It can be seen that the right side of the equation is x 2 + 1 · e x. From this, the LNDE is found through y ~ \u003d e a x Q n (x) x γ, where Q n (x), which is a polynomial of the second degree, where α \u003d 1 and r \u003d 0, because the characteristic equation has no root equal to 1. From this 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 by the equality y ~ "" - 2 y ~ "\u003d (x 2 + 1) · e x.
Got that
y ~ "\u003d ex A x 2 + B x + C" \u003d ex A x 2 + B x + C + ex 2 A x + B \u003d \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 are equated and we get a system of linear equations. From here we 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 is 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 particular solution of the LNDE, and y \u003d y 0 + y \u003d C 1 e 2 x - ex · x 2 + 3 - the general solution for the inhomogeneous second order differential equation.
When the function is written as f (x) \u003d A 1 cos (β x) + B 1 sin β x, and A 1 and IN 1are numbers, then an equation of the form y ~ \u003d A cos β x + B sin β x x γ, where A and B are considered as indefinite 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 the coefficients is carried out according to the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 3
Find the general solution of a differential equation of the form y "" + 4 y \u003d cos (2 x) + 3 sin (2 x).
Decision
Before writing the characteristic equation, we find y 0. Then
k 2 + 4 \u003d 0 k 2 \u003d - 4 k 1 \u003d 2 i, k 2 \u003d - 2 i
We have a pair of complex conjugate roots. Let's transform and get:
y 0 \u003d e 0 (C 1 cos (2 x) + C 2 sin (2 x)) \u003d C 1 cos 2 x + C 2 sin (2 x)
The roots from the characteristic equation are considered to be the conjugate pair ± 2 i, then f (x) \u003d cos (2 x) + 3 sin (2 x). Hence, it can be seen that the search for y ~ will be performed 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 sought from an equality of the form y ~ "" + 4 y ~ \u003d cos (2 x) + 3 sin (2 x).
Let's transform:
y ~ "\u003d ((A cos (2 x) + B sin (2 x) x)" \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 is clear 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 the sines and cosines. We get a system of the form:
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 original second-order LDE with constant coefficients is
y \u003d y 0 + y ~ \u003d \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 complex conjugate pairs of roots belonging to the characteristic equation equal to α ± i β, where P n (x), Q k (x), L m (x) and N m (x)are polynomials of degree n, k, m, m, where m \u003d m a x (n, k)... Finding coefficients L m (x) and N m (x) is produced proceeding from the equality y ~ "" + p · y ~ "+ q · y ~ \u003d f (x).
Example 4
Find the General Solution y "" + 3 y "+ 2 y \u003d - e 3 x · ((38 x + 45) sin (5 x) + (8 x - 5) cos (5 x)).
Decision
By condition, it is 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 find y 0, having previously written down 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
Got the roots to be valid and distinct. Hence y 0 \u003d C 1 e x + C 2 e 2 x. Next, it is necessary to seek a general solution based on the inhomogeneous equation y ~ of the form
y ~ \u003d e α x (L m (x) sin (β x) + N m (x) cos (β x) x γ \u003d \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 related to the characteristic equation with α ± i β \u003d 3 ± 5 · i. We find these coefficients from the obtained equality:
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 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)) \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 the coefficients, we obtain a system of the form
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 general solution of the given 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 for solving LDNU
Definition 1Any other kind of function f (x) for the solution provides for the observance of the solution algorithm:
- finding a general solution to the corresponding linear homogeneous equation, where y 0 \u003d C 1 ⋅ y 1 + C 2 ⋅ y 2, where y 1 and y 2are linearly independent particular solutions of the LODE, C 1 and C 2are considered arbitrary constants;
- adoption as a general decision of the LNDE y \u003d C 1 (x) ⋅ y 1 + C 2 (x) ⋅ y 2;
- definition of the derivatives of a function in terms of a system of the form 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 the functions C 1 (x) and C 2 (x) through integration.
Example 5
Find the General Solution for y "" + 36 y \u003d 24 sin (6 x) - 12 cos (6 x) + 36 e 6 x.
Decision
We proceed to writing the characteristic equation, having previously written down y 0, y "" + 36 y \u003d 0. Let's write down and solve:
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 record of the general solution of the given equation will receive the form y \u003d C 1 (x) · cos (6 x) + C 2 (x) · sin (6 x). It is necessary to go to the definition of derivatives of functions C 1 (x) and C 2 (x) according to the system with equations:
C 1 "(x) cos (6 x) + C 2" (x) sin (6 x) \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 \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 method. Then we write:
C 1 "(x) \u003d - 4 sin 2 (6 x) + 2 sin (6 x) cos (6 x) - 6 e 6 x sin (6 x) C 2" (x) \u003d 4 sin (6 x) cos (6 x) - 2 cos 2 (6 x) + 6 e 6 x cos (6 x)
Each of the equations 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
From this it follows that the general solution will be:
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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