Method of variation of arbitrary constants for inhomogeneous equations. Method of Variation of Arbitrary Constants
Consider now the linear inhomogeneous equation
. (2)
Let y 1 ,y 2 ,.., y n be the fundamental system of solutions, and be the general solution of the corresponding homogeneous equation L(y)=0 . Similarly to the case of first-order equations, we will seek a solution to Eq. (2) in the form
. (3)
Let us verify that a solution in this form exists. To do this, we substitute the function into the equation. To substitute this function into the equation, we find its derivatives. The first derivative is 
. (4)
When calculating the second derivative, four terms appear on the right side of (4), when calculating the third derivative, eight terms appear, and so on. Therefore, for the convenience of further calculations, the first term in (4) is assumed to be equal to zero. With this in mind, the second derivative is equal to 
. (5)
For the same reasons as before, in (5) we also set the first term equal to zero. Finally, the nth derivative is 
. (6)
Substituting the obtained values of the derivatives into the original equation, we have 
. (7)
The second term in (7) is equal to zero, since the functions y j , j=1,2,..,n, are solutions of the corresponding homogeneous equation L(y)=0. Combining with the previous one, we obtain a system of algebraic equations for finding the functions C" j (x) 
(8)
The determinant of this system is the Wronsky determinant of the fundamental system of solutions y 1 ,y 2 ,..,y n of the corresponding homogeneous equation L(y)=0 and therefore is not equal to zero. Therefore, there is a unique solution to system (8). Having found it, we obtain the functions C "j (x), j=1,2,…,n, and, consequently, C j (x), j=1,2,…,n Substituting these values into (3), we obtain the solution of the linear inhomogeneous equation.
The described method is called the method of variation of an arbitrary constant or the Lagrange method.
Example #1. Let's find the general solution of the equation y "" + 4y" + 3y \u003d 9e -3 x. Consider the corresponding homogeneous equation y "" + 4y" + 3y \u003d 0. The roots of its characteristic equation r 2 + 4r + 3 \u003d 0 are equal to -1 and - 3. Therefore, the fundamental system of solutions of a homogeneous equation consists of the functions y 1 = e - x and y 2 = e -3 x. We are looking for a solution to an inhomogeneous equation in the form y \u003d C 1 (x)e - x + C 2 (x)e -3 x. To find the derivatives C " 1 , C" 2 we compose a system of equations (8)
C′ 1 ·e -x +C′ 2 ·e -3x =0
-C′ 1 e -x -3C′ 2 e -3x =9e -3x
solving which, we find , Integrating the obtained functions, we have 
Finally we get
Example #2. Solve linear differential equations of the second order with constant coefficients by the method of variation of arbitrary constants: 
y(0) =1 + 3ln3
y'(0) = 10ln3
Solution:
This differential equation belongs to linear differential equations with constant coefficients.
We will seek the solution of the equation in the form y = e rx . To do this, we compose the characteristic equation of a linear homogeneous differential equation with constant coefficients:
r 2 -6 r + 8 = 0
D = (-6) 2 - 4 1 8 = 4
The roots of the characteristic equation: r 1 = 4, r 2 = 2
Therefore, the fundamental system of solutions is the functions: y 1 =e 4x , y 2 =e 2x
The general solution of the homogeneous equation has the form: y =C 1 e 4x +C 2 e 2x
Search for a particular solution by the method of variation of an arbitrary constant.
To find the derivatives of C "i, we compose a system of equations:
C′ 1 e 4x +C′ 2 e 2x =0
C′ 1 (4e 4x) + C′ 2 (2e 2x) = 4/(2+e -2x)
Express C" 1 from the first equation:
C" 1 \u003d -c 2 e -2x
and substitute in the second. As a result, we get:
C" 1 \u003d 2 / (e 2x + 2e 4x)
C" 2 \u003d -2e 2x / (e 2x + 2e 4x)
We integrate the obtained functions C" i:
C 1 = 2ln(e -2x +2) - e -2x + C * 1
C 2 = ln(2e 2x +1) – 2x+ C * 2
Since y \u003d C 1 e 4x + C 2 e 2x, then we write the resulting expressions in the form:
C 1 = (2ln(e -2x +2) - e -2x + C * 1) e 4x = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x
C 2 = (ln(2e 2x +1) – 2x+ C * 2)e 2x = e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
Thus, the general solution of the differential equation has the form:
y = 2 e 4x ln(e -2x +2) - e 2x + C * 1 e 4x + e 2x ln(2e 2x +1) – 2x e 2x + C * 2 e 2x
or
y = 2 e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + C * 1 e 4x + C * 2 e 2x
We find a particular solution under the condition:
y(0) =1 + 3ln3
y'(0) = 10ln3
Substituting x = 0 into the found equation, we get:
y(0) = 2 ln(3) - 1 + ln(3) + C * 1 + C * 2 = 3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
We find the first derivative of the obtained general solution:
y’ = 2e 2x (2C 1 e 2x + C 2 -2x +4 e 2x ln(e -2x +2)+ ln(2e 2x +1)-2)
Substituting x = 0, we get:
y'(0) = 2(2C 1 + C 2 +4 ln(3)+ ln(3)-2) = 4C 1 + 2C 2 +10 ln(3) -4 = 10ln3
We get a system of two equations:
3 ln(3) - 1 + C * 1 + C * 2 = 1 + 3ln3
4C 1 + 2C 2 +10ln(3) -4 = 10ln3
or
C * 1 + C * 2 = 2
4C1 + 2C2 = 4
or
C * 1 + C * 2 = 2
2C1 + C2 = 2
From: C 1 = 0, C * 2 = 2
A particular solution will be written as:
y = 2e 4x ln(e -2x +2) - e 2x + e 2x ln(2e 2x +1) – 2x e 2x + 2 e 2x
Theoretical minimumIn the theory of differential equations, there is a method that claims to have a sufficiently high degree of universality for this theory.
We are talking about the method of variation of an arbitrary constant, applicable to the solution of various classes of differential equations and their
systems. This is exactly the case when the theory - if you take the proof of the statements out of brackets - is minimal, but allows you to achieve
significant results, so the main focus will be on examples.The general idea of the method is quite simple to formulate. Let the given equation (system of equations) be difficult to solve or even incomprehensible,
how to solve it. However, it can be seen that when some terms are excluded from the equation, it is solved. Then they solve just such a simplified
equation (system), get a solution containing a certain number of arbitrary constants - depending on the order of the equation (the number
equations in the system). Then it is assumed that the constants in the found solution are not really constants, the found solution
is substituted into the original equation (system), a differential equation (or system of equations) is obtained to determine the "constants".
There is a certain specificity in applying the method of variation of an arbitrary constant to different problems, but these are already details that will be
shown with examples.Let us separately consider the solution of linear inhomogeneous equations of higher orders, i.e. equations of the form
.
The general solution of a linear inhomogeneous equation is the sum of the general solution of the corresponding homogeneous equation and the particular solution
given equation. Let us assume that the general solution of the homogeneous equation has already been found, namely, the fundamental system of solutions (FSR) has been constructed
. Then the general solution of the homogeneous equation is .
It is necessary to find any particular solution of the inhomogeneous equation. For this, constants are considered to be dependent on the variable.
Next, you need to solve the system of equations.
The theory guarantees that this system of algebraic equations with respect to derivatives of functions has a unique solution.
When finding the functions themselves, the integration constants do not appear: after all, any one solution is sought.In the case of solving systems of linear inhomogeneous equations of the first order of the form
the algorithm remains almost unchanged. First you need to find the FSR of the corresponding homogeneous system of equations, compose the fundamental matrix
system , the columns of which are the elements of the FSR. Next, the equation.
Solving the system, we determine the functions , thus finding a particular solution to the original system
(the fundamental matrix is multiplied by the found feature column).
We add it to the general solution of the corresponding system of homogeneous equations, which is built on the basis of the FSR already found.
The general solution of the original system is obtained.Examples.
Example 1 Linear inhomogeneous equations of the first order.
Let us consider the corresponding homogeneous equation (we denote the required function by ):
.
This equation is easily solved by separation of variables:
.
Now we represent the solution of the original equation in the form, where the function is yet to be found.
We substitute this type of solution into the original equation:
.
As you can see, the second and third terms on the left side cancel each other - this is a characteristic feature of the method of variation of an arbitrary constant.
Here already - indeed, an arbitrary constant. Thus,
.Example 2 Bernoulli equation.
We act similarly to the first example - we solve the equation
method of separation of variables. It will turn out , so we are looking for the solution of the original equation in the form
.
Substitute this function into the original equation:
.
And again there are cuts:
.
Here you need to remember to make sure that when dividing by, the solution is not lost. And the case corresponds to the solution of the original
equations. Let's remember him. So,.
Let's write .
This is the solution. When writing the answer, you should also indicate the solution found earlier, since it does not correspond to any final value
constants .Example 3 Linear inhomogeneous equations of higher orders.
We note right away that this equation can be solved more simply, but it is convenient to show the method on it. Although some advantages
the method of variation of an arbitrary constant also has it in this example.
So, you need to start with the FSR of the corresponding homogeneous equation. Recall that in order to find the FSR, the characteristic
the equation
.
Thus, the general solution of the homogeneous equation
.
The constants included here are to be varied. Compiling a system
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 the remote control, 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 he is 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 (constant) is also one.
2) The method of variation of arbitrary constants is used to solve some linear inhomogeneous equations of the second order. Here, two constants (constants) vary.
It is logical to assume that the lesson will consist of two paragraphs .... I wrote this proposal, and for about 10 minutes I was painfully thinking what other smart crap to add for a smooth transition to practical examples. But for some reason, there are no thoughts after the holidays, although it seems that I did not abuse anything. So let's jump right into the first paragraph.
Arbitrary Constant Variation Method
for a linear inhomogeneous first-order equation
Before considering the method of variation of an arbitrary constant, it is desirable to be familiar with the article Linear differential equations of the first order. In that lesson, we practiced first way to solve inhomogeneous 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 way to solve– method of variation of an arbitrary constant. I will give only 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 replacement method.
Example 1
(Diffur from Example No. 2 of the lesson Linear inhomogeneous DE of the 1st order)
Solution: This equation is linear inhomogeneous and has a familiar form: 
The first step is to solve a simpler equation: 
That is, we stupidly reset the right side - instead we write zero.
The equation I'll call 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: 
is the general solution of the auxiliary equation .
On 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 initial inhomogeneous equation Let's replace:
Substitute and 
into the equation :
control moment - 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 is obtained. Separate variables and integrate.
What a blessing, the exponents are shrinking too:
We add a “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 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 something more complicated, I will also comment on the second example:
Example 2
Find the general solution of the differential equation 
(Diffur from Example No. 8 of lesson Linear inhomogeneous DE of the 1st order)
Solution: We bring the equation to the form :
Set the right side to zero and solve the auxiliary equation:
General solution of the auxiliary equation:
In the inhomogeneous equation, we will make the substitution:
According to the product differentiation rule: 
Substitute and into the original inhomogeneous equation :
The two terms on the left side cancel out, which means we are on the right track: 
We integrate by parts. A tasty letter from the formula for integration by parts is already involved in the solution, so we use, for example, the letters "a" and "be":
Now let's look at the replacement:
Answer: common decision:
And one example for self solution:
Example 3
Find a particular solution of the differential equation corresponding to the given initial condition.
,
(Diffur from Lesson 4 Example Linear inhomogeneous DE of the 1st order)
Solution:
This DE is linear inhomogeneous. We use the method of variation of arbitrary constants. Let's solve the auxiliary equation:
We separate the variables and integrate: 
Common decision: 
In the inhomogeneous equation, we will make the substitution: 
Let's do the substitution: 
So the general solution is: 
Find a particular solution corresponding to the given initial condition: 
Answer: private solution:
The solution at the end of the lesson can serve as an approximate model for finishing the assignment.
Method of Variation of Arbitrary Constants
for a linear inhomogeneous second order equation
with constant coefficients
One 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, and understandable. And beautiful.
To master the method, it is desirable to be able to solve inhomogeneous equations of the second order by selecting a particular solution according to the form of the right side. This method is discussed in detail in the article. Inhomogeneous DE of the 2nd order. We recall that a 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, a fraction, logarithm, tangent? In such a situation, the method of variation of constants comes to the rescue.
Example 4
Find the general solution of 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.
Nothing portends a thunderstorm, the beginning of the solution is quite ordinary:
Let's find common decision relevant homogeneous equations: 
We 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 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 the inhomogeneous We will look for equations in the form:
Where - yet unknown functions.
It looks like a garbage dump, but now we'll 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 "games" come from? The stork brings them. We look at the previously obtained general solution and write:
Let's find derivatives:
Dealt with the left side. What's on the right?
is the right side of the original equation, in this case:
The coefficient is the coefficient at the second derivative:
In practice, almost always, and our example is no exception.
Everything cleared up, now you can create a system: 
The system is usually solved according to Cramer's formulas using the standard algorithm. The only difference is that instead of numbers we have functions.
Find the main determinant of the system: 
If you forgot how the “two by two” determinant is revealed, refer to the lesson How to calculate the determinant? The link leads to the board of shame =)
So: , so the system has a unique solution.
We find the derivative: 
But that's not all, so far we've only found the derivative.
The function itself is restored by integration:
Let's look at the second function: 
Here we add a "normal" constant
At the final stage of the solution, we recall in what form we were looking for the general solution of the inhomogeneous equation? In such:
The features you need have just been found! 
It remains to perform the substitution and write down the answer:
Answer: common decision:
In principle, the answer could open the brackets.
A full check of the answer is performed according to the standard scheme, which was considered in the lesson. Inhomogeneous DE of the 2nd order. But the verification will not be easy, since we have to find rather heavy derivatives and carry out a cumbersome substitution. This is a nasty feature when you're solving diffs like this.
Example 5
Solve the differential equation by the method of variation of arbitrary constants
This is a do-it-yourself example. In fact, the right side is also a fraction. We recall the trigonometric formula, by the way, it will need to be applied along the way.
The method of variation of arbitrary constants is the most universal method. They can solve any equation that can be solved the method of selecting a particular solution according to the form of the right side. The question arises, why not use the method of variation of arbitrary constants there as well? The answer is obvious: the selection of a particular solution, which was considered in the lesson Inhomogeneous equations of the second order, significantly speeds up the solution and reduces the notation - no messing around with determinants and integrals.
Consider two examples with Cauchy problem.
Example 6
Find a particular solution of the differential equation corresponding to given initial conditions
, 
Solution: Again a fraction and an exponent in interesting place.
We use the method of variation of arbitrary constants.
Let's find common decision relevant homogeneous equations: 
– different real roots are obtained, so the general solution is:
The general solution of the inhomogeneous we are looking for equations in the form: , where - yet unknown functions.
Let's create a system:
In this case:
,
Finding derivatives: 
, 
Thus: 
We solve the system using Cramer's formulas:
, so the system has a unique solution.
We restore the function by integration:
Used here method of bringing a function under a differential sign.
We restore the second function by integration: 
Such an integral is solved variable substitution method:
From the replacement itself, we express:
Thus: 
This integral can be found full square selection method, but in examples with diffurs, I prefer to expand the fraction method of uncertain coefficients:
Both functions found:
As a result, the general solution of the inhomogeneous equation is:
Find a particular solution that satisfies the initial conditions .
Technically, the search for a solution is carried out in a standard way, which was discussed in the article. Inhomogeneous Second Order Differential Equations.
Hold on, now we will find the derivative of the found general solution:
Here is such a disgrace. It is not necessary to simplify it, it is easier to immediately compose a system of equations. According to the initial conditions :
Substitute the found values of the constants 
into a general solution:
In the answer, the logarithms can be packed a little.
Answer: private solution: 
As you can see, difficulties can arise in integrals and derivatives, but not in the algorithm of the method of variation of arbitrary constants. It was not I who intimidated you, this is all a collection of Kuznetsov!
To relax, a final, simpler, self-solving example:
Example 7
Solve the Cauchy problem
, 
The example is simple, but creative, when you make a system, look at it carefully before deciding ;-),
As a result, the general solution is:
Find a particular solution corresponding to the initial conditions .
We substitute the found values of the constants into the general solution:
Answer: private solution:
Lecture 44. Linear inhomogeneous equations of the second order. Method of variation of arbitrary constants. Linear inhomogeneous equations of the second order with constant coefficients. (special right side).
Social transformations. State and Church.
The social policy of the Bolsheviks was largely dictated by their class approach. By a decree of November 10, 1917, the estate system was abolished, pre-revolutionary ranks, titles and awards were abolished. The election of judges has been established; the secularization of civil states was carried out. Established free education and medical care (decree of October 31, 1918). Women were equalized in rights with men (decrees of December 16 and 18, 1917). The decree on marriage introduced the institution of civil marriage.
By a decree of the Council of People's Commissars of January 20, 1918, the church was separated from the state and from the education system. Much of the church property was confiscated. Patriarch Tikhon of Moscow and All Rus' (elected November 5, 1917) on January 19, 1918, anathematized Soviet power and called for a fight against the Bolsheviks.
Consider a linear inhomogeneous second-order equation
The structure of the general solution of such an equation is determined by the following theorem:
Theorem 1. The general solution of the inhomogeneous equation (1) is represented as the sum of some particular solution of this equation and the general solution of the corresponding homogeneous equation
Proof. We need to prove that the sum
is the general solution of equation (1). Let us first prove that function (3) is a solution of equation (1).
Substituting the sum into equation (1) instead of at, will have
Since there is a solution to equation (2), the expression in the first brackets is identically equal to zero. Since there is a solution to equation (1), the expression in the second brackets is equal to f(x). Therefore, equality (4) is an identity. Thus, the first part of the theorem is proved.
Let us prove the second assertion: expression (3) is general solution of equation (1). We must prove that the arbitrary constants included in this expression can be chosen so that the initial conditions are satisfied:
whatever the numbers x 0 , y 0 and (if only x 0 was taken from the area where the functions a 1 , a 2 And f(x) continuous).
Noticing that it is possible to represent in the form . Then, based on conditions (5), we have
Let's solve this system and find From 1 And From 2. Let's rewrite the system as:
Note that the determinant of this system is the Wronsky determinant for the functions 1 And at 2 at the point x=x 0. Since these functions are linearly independent by assumption, the Wronsky determinant is not equal to zero; hence system (6) has a definite solution From 1 And From 2, i.e. there are such values From 1 And From 2, for which formula (3) determines the solution of equation (1) that satisfies the given initial conditions. Q.E.D.
Let us turn to the general method for finding particular solutions of an inhomogeneous equation.
Let us write the general solution of the homogeneous equation (2)
We will look for a particular solution of the inhomogeneous equation (1) in the form (7), considering From 1 And From 2 as some as yet unknown features from X.
Let us differentiate equality (7):
We select the desired functions From 1 And From 2 so that the equality
If this additional condition is taken into account, then the first derivative takes the form
Now differentiating this expression, we find:
Substituting into equation (1), we obtain
The expressions in the first two brackets vanish because y 1 And y2 are solutions of a homogeneous equation. Therefore, the last equality takes the form
Thus, function (7) will be a solution to the inhomogeneous equation (1) if the functions From 1 And From 2 satisfy equations (8) and (9). Let us compose a system of equations from equations (8) and (9).
Since the determinant of this system is the Vronsky determinant for linearly independent solutions y 1 And y2 equation (2), then it is not equal to zero. Therefore, solving the system, we will find both certain functions of X:
Solving this system, we find , whence, as a result of integration, we obtain . Next, we substitute the found functions into the formula , we obtain the general solution of the inhomogeneous equation , where are arbitrary constants.
Method of Variation of Arbitrary Constants
Method of variation of arbitrary constants for constructing a solution to a linear inhomogeneous differential equation
a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = f(t)
consists in changing arbitrary constants c k in the general decision
z(t) = c 1 z 1 (t) + c 2 z 2 (t) + ... + c n z n (t)
corresponding homogeneous equation
a n (t)z (n) (t) + a n − 1 (t)z (n − 1) (t) + ... + a 1 (t)z"(t) + a 0 (t)z(t) = 0
to helper functions c k (t) , whose derivatives satisfy the linear algebraic system
The determinant of system (1) is the Wronskian of functions z 1 ,z 2 ,...,z n , which ensures its unique solvability with respect to .
If are antiderivatives for taken at fixed values of the constants of integration, then the function
is a solution to the original linear inhomogeneous differential equation. Integration of an inhomogeneous equation in the presence of a general solution of the corresponding homogeneous equation is thus reduced to quadratures.
Method of variation of arbitrary constants for constructing solutions to a system of linear differential equations in vector normal form
consists in constructing a particular solution (1) in the form
Where Z(t) is the basis of solutions of the corresponding homogeneous equation, written as a matrix, and the vector function , which replaced the vector of arbitrary constants, is defined by the relation . The desired particular solution (with zero initial values at t = t 0 has the form
For a system with constant coefficients, the last expression is simplified:
Matrix Z(t)Z− 1 (τ) called Cauchy matrix operator L = A(t) .
