At the lecture, LNDEs are studied - linear inhomogeneous differential equations. The structure of the general solution is considered, the solution of LPDE using the method of variation of arbitrary constants, the solution of LPDE with constant coefficients
and the right side of a special type. The issues under consideration are used in the study of forced oscillations in physics, electrical engineering and electronics, and the theory of automatic control.
1. Structure of the general solution of a linear inhomogeneous differential equation of 2nd order.
Let us first consider a linear inhomogeneous equation of arbitrary order:
Taking into account the notation, we 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 a certain domain is the sum of any of its solutions and the general solution of the corresponding linear homogeneous differential equation. Proof.
Let Y be some solution to an inhomogeneous equation.
Then, when substituting this solution into the original equation, we obtain the identity: 
- Let fundamental system 
solutions to a linear homogeneous equation . Then common decision
homogeneous equation can be written as:
In particular, for a linear inhomogeneous differential equation of 2nd order, the structure of the general solution has the form: 
Where 
is the fundamental system of solutions to 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 a general solution to the corresponding homogeneous equation and somehow find one particular solution inhomogeneous equation
. Usually it is found by selection. We will consider methods for selecting a private solution in the following questions.
In practice, it is convenient to use the method of varying arbitrary constants.
To do this, first find a general solution to the corresponding homogeneous equation in the form:
Then, putting the coefficients C i functions from X, a solution to the inhomogeneous equation is sought:
It can be proven that to find functions C i (x) we need to solve the system of equations:
Example. Solve the equation 
Solving a linear homogeneous equation 
The solution to the inhomogeneous equation will have the form:
Let's create 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 into 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 because Finding the fundamental system of solutions to the corresponding homogeneous equation can be quite a difficult task; this method is mainly used for inhomogeneous equations with constant coefficients.
3. Equations with the right side of a special form
It seems possible to imagine the type of a particular solution depending on the type 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) - polynomial of the same degree as P(x) , nose uncertain coefficients, A 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 us solve the corresponding homogeneous equation: 
Now let's find a particular solution to the original inhomogeneous equation.
Let's compare the right side of the equation with the form of the right side discussed above.
We look for a particular solution in the form: 
, Where
Those. 
Now let's determine the unknown coefficients A And IN.
Let us substitute a particular solution in general form into the original inhomogeneous differential equation.
Total, private solution: 
Then the general solution of a linear inhomogeneous differential equation is:
II.
Here The right side of the linear inhomogeneous differential equation has the form: 1 R(X) The right side of the linear inhomogeneous differential equation has the form: 2 R And m– polynomials of degree m 2 1 and
respectively.
Then a particular solution to the inhomogeneous equation will have the form: r where is the number 
shows how many times a number Q 1
(x)
And
Q 2
(x)
is the root of the characteristic equation for the corresponding homogeneous equation, and m– polynomials of degree not higher than m, Where m 1
- the largest of the degrees m 2
.
And
Summary table of types of private solutions
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Right side of the differential equation |
characteristic equation |
Types of private |
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1. The number is not the root of the characteristic equation |
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2. Number is 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 multiplicity equation 
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 type 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 is: 
, then a particular solution to this equation will be 
Where at 1
(X) at 2
– particular solutions of auxiliary equations
(X) 
To illustrate, let's solve the above example in a different way.
Example. Solve the equation 
Let us 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: I.e. 
Total: 
Those. the required particular solution has the form: 
General solution of a non-homogeneous differential equation:
Let's look at examples of the application of the described methods.
Example 1.. Solve the equation 
Let us compose a characteristic equation for the corresponding linear homogeneous differential equation:
Now we find a particular solution to the inhomogeneous equation in the form:
Let's use the method of indefinite coefficients.
Substituting into the original equation, we get:
A particular solution has the form: 
General solution of a linear inhomogeneous equation: 
Example. Solve the equation 
Characteristic equation:
General solution of the homogeneous equation: 
Particular solution of the inhomogeneous equation: 
.
We 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 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 characteristic equation corresponding to LOD-2, 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 left side 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 linear equations relative to 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) $.
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Inhomogeneous second order differential equations with constant coefficients |
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Structure of the general solution A linear inhomogeneous equation of this type has the form: Where p, q− constant numbers (which can be either real or complex). For each such equation we can write the corresponding:
homogeneous equation Theorem : The general solution of an inhomogeneous equation is the sum of the general solution 0 (x y : The general solution of an inhomogeneous equation is the sum of the general solution 1 (x) of the corresponding homogeneous equation and particular solution
) inhomogeneous equation: Below we will consider two ways to solve inhomogeneous differential equations. Method of variation of constants : The general solution of an inhomogeneous equation is the sum of the general solution If the general solution 0 of the associated homogeneous equation is known, then the general solution to the inhomogeneous equation can be found using constant variation method . Let the general solution of a homogeneous second-order differential equation have the form: C Instead of permanent C 1 and C 1 (x 2 we will consider auxiliary functions C 2 (x) And ). f(x We will look for these functions such that the solution C 1 (x 2 we will consider auxiliary functions C 2 (x satisfied the inhomogeneous equation with the right-hand side
). Unknown functions ) are determined from a system of two equations: f(x Uncertain coefficient method Right part) of an inhomogeneous differential equation is often a polynomial, exponential or trigonometric function, or some combination of these functions. In this case, it is more convenient to search for a solution using method of uncertain coefficients. Let us emphasize that  this method α 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 V x exponential function coincides with the root of the characteristic equation, then the particular solution will contain an additional factor exponential function s α , Where − multiplicity of root in the characteristic equation. In case 2, if the number xα + βi coincides with the root of the characteristic equation, then the expression for the particular solution will contain an additional factor . 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 then a particular solution of the differential equation will also be the sum of partial solutions constructed separately for each term on the right side. |
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Example 1 |
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Solve differential equation y"" + y= sin(2 x). Solution. First we solve the corresponding homogeneous equation y"" + y= 0.V in this case the roots of the characteristic equation are purely imaginary: Consequently, the general solution of the homogeneous equation is given by the expression Let's return again to the inhomogeneous equation. We will look for its solution in the form using the method of variation of constants. Functions C 1 (x 2 we will consider auxiliary functions C 2 (x) can be found from next system equations:
Let's express the derivative C 1 " (x) from the first equation:
Substituting into the second equation, we find the derivative C 2 " (x):
It follows that Integrating expressions for derivatives C 1 " (x 2 we will consider auxiliary functions C 2 " (x), we get: Where A 1 , A 2 – constants of integration. Now let’s substitute the found functions C 1 (x 2 we will consider auxiliary functions C 2 (x) into the formula for : The general solution of an inhomogeneous equation is the sum of the general solution 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" −6: The general solution of an inhomogeneous equation is the sum of the general solution = 36x. Solution. Let's use the method of indefinite coefficients. The right side of the given equation is f(x)linear function= ax + b .
Therefore, we will look for a particular solution in the form
The derivatives are equal: x Substituting this into the differential equation, we get: x The last equation is an identity, that is, it is valid for all , therefore we equate the coefficients of terms with the same degrees A = −6, on the left and right side: From the resulting system we find: B = −1. As a result, the particular solution is written in the form Now let's find the general solution of the homogeneous differential equation. Let us calculate the roots of the auxiliary characteristic equation: |
Therefore, the general solution of the corresponding homogeneous equation has the form:
So, the general solution of the original inhomogeneous equation is expressed by the formula
General integral of the DE.
Solve differential equation
But the funniest thing is that the answer is already known: , more precisely, we must also add a constant: The general integral is a solution to the differential equation. Method of variation of arbitrary constants. Examples of solutions. And if you are already finishing, please discard the possible preconception that the method is difficult. Because it's simple.
In what cases is the method of variation of arbitrary constants used?
1) The method of variation of an arbitrary constant can be used to solve linear inhomogeneous DE of the 1st order. Since the equation is of the first order, then the constant is also one.
2) The method of variation of arbitrary constants is used to solve some linear inhomogeneous second order equations. Here two constants vary.
It is logical to assume that the lesson will consist of two paragraphs... So I wrote this sentence, and for about 10 minutes I was painfully thinking about what other clever crap I could add for a smooth transition to practical examples. But for some reason I don’t have any thoughts after the holidays, although I don’t seem to have abused anything. Therefore, let's get straight to the first paragraph.
Method of variation of an arbitrary constant for a first order linear inhomogeneous 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 inhomogeneous 1st order DE. This first solution, I remind you, is called replacement method or Bernoulli method(not to be confused with Bernoulli's equation!!!)
Now we will look second solution– method of variation of an arbitrary constant. I will give just three examples, and I will take them from the above-mentioned 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 frequently than the replacement method.
Example 1
Find the general solution of the differential equation (Diffour from Example No. 2 of the lesson Linear inhomogeneous differential equations of the 1st order)
Solution: 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 reset the right side to zero - write zero instead. I will call the equation auxiliary equation.
In this example, you need to solve the following auxiliary equation:
Before us separable equation, the solution of which (I hope) is no longer difficult for you:
Thus: – general solution of the auxiliary equation.
On the second step we will replace some constant for now unknown function that depends on "x":
Hence the name of the method - we vary the constant. Alternatively, the constant could be some function that we now have to find.
IN original in the inhomogeneous equation we make the replacement:
Let's substitute into the equation:
Control point – the two terms on the left side cancel. If this does not happen, you should look for the error above.
As a result of the replacement, an equation with separable variables was obtained. We separate the variables and integrate.
What a blessing, the exponents also cancel:
We add a “normal” constant to the found function:
At the final stage, we remember our replacement:
The function has just been found!
So the general solution is:
Answer: common decision: 
If you print out the 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 of the differential equation (Diffour from Example No. 8 of lesson Linear inhomogeneous differential equations of the 1st order)
Solution: Let's bring the equation to the form:
Let's reset the right-hand side and solve the auxiliary equation:
We separate the variables and integrate: The general solution to the auxiliary equation:
In the inhomogeneous equation we make the replacement:
According to the product differentiation rule:
Let us substitute into the original inhomogeneous equation:
The two terms on the left side cancel, which means we are on the right track: 
Let's integrate by parts. The tasty letter from the integration by parts formula is already involved in the solution, so we use, for example, the letters “a” and “be”:
Eventually:
Now let's remember the replacement:
Answer: common decision:
Method of variation of arbitrary constants for a linear inhomogeneous second order equation with constant coefficients
I have often heard the opinion that the method of varying arbitrary constants for a second-order equation is not an easy thing. But I assume the following: most likely, the method seems difficult to many because it does not occur so often. But in reality there are no particular difficulties - the course of the decision is clear, transparent, and understandable. And beautiful.
To master the method, it is desirable to be able to solve inhomogeneous second-order equations by selecting a particular solution based on the form of the right-hand side. This method discussed in detail in the article Inhomogeneous 2nd order DEs. We recall that a second-order linear inhomogeneous equation with constant coefficients has the form:
The selection method, which was discussed in the above lesson, works only in a limited number of cases when the right side contains polynomials, exponentials, sines, and cosines. But what to do when on the right, for example, is a fraction, logarithm, tangent? In such a situation, the method of variation of constants comes to the rescue.
Example 4
Find the general solution to a second order differential equation 
Solution: 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.
There are no signs of a thunderstorm; the beginning of the solution is completely ordinary:
We'll find common decision appropriate 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 parentheses, then open them.
Now we do almost the same trick as for the first-order equation: we vary the constants, replacing them with unknown functions. That is, general solution of inhomogeneous we will look for equations in the form:
Where - for now unknown functions.
Looks like a landfill household waste, but now we’ll sort everything out.
The unknowns are the derivatives of the functions. Our goal is to find derivatives, and the found derivatives must satisfy both the first and second equations of the system.
Where do the “Greeks” come from? The stork brings them. We look at the general solution obtained earlier and write:
Let's find the derivatives:
The left parts have been dealt with. What's on the right?
- this is the right side original equation, in this case:
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 1A general solution, located on the interval x, of an inhomogeneous differential equation of the form y (n) + f n - 1 (x) · y (n - 1) + . . . + f 0 (x) · y = f (x) with continuous integration coefficients on the x interval f 0 (x) , f 1 (x) , . . . , f n - 1 (x) and continuous function f (x) is equal to the sum of the general solution y 0, which corresponds to the 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 equation general view 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 transform:
y ~ " = ((A cos (2 x) + B sin (2 x) x) " = = (- 2 A sin (2 x) + 2 B cos (2 x)) x + A cos (2 x) + B sin (2 x) y ~ "" = ((- 2 A sin (2 x) + 2 B cos (2 x)) x + A cos (2 x) + B sin (2 x)) " = = (- 4 A cos (2 x) - 4 B sin (2 x)) x - 2 A sin (2 x) + 2 B cos (2 x) - - 2 A sin (2 x) + 2 B cos (2 x) = = (- 4 A cos (2 x) - 4 B sin (2 x)) x - 4 A sin (2 x) + 4 B cos (2 x)
Then it is clear that
y ~ "" + 4 y ~ = cos (2 x) + 3 sin (2 x) ⇔ (- 4 A cos (2 x) - 4 B sin (2 x)) x - 4 A sin (2 x) + 4 B cos (2 x) + + 4 (A cos (2 x) + B sin (2 x)) x = cos (2 x) + 3 sin (2 x) ⇔ - 4 A sin (2 x) + 4 B cos (2 x) = cos (2 x) + 3 sin (2 x)
It is necessary to equate the coefficients of 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 1Any 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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