Differential equations method of variation of constant. Method of variation of an arbitrary constant for solving linear inhomogeneous equations
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 remote control, i.e. If you are a teapot, I recommend starting with the first lesson: First order differential equations. 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 only 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
(Diffur 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 and write zero instead.
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: 
– 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 inhomogeneous equation let's make a replacement:
Let's substitute and 
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 to the differential equation 
(Diffur from Example No. 8 of lesson Linear inhomogeneous differential equations of the 1st order)
Solution: Let us reduce the equation to the form :
Let's reset the right-hand side and solve the auxiliary equation:
General solution to the auxiliary equation:
In the inhomogeneous equation we make the replacement:
According to the product differentiation rule: 
Let's substitute and 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”:
Now let's remember the replacement:
Answer: common decision:
And one example for an independent solution:
Example 3
Find a particular solution to the differential equation corresponding to the given initial condition.
,
(Diffur from Example No. 4 of the lesson Linear inhomogeneous differential equations 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 make the replacement: 
Let's perform the substitution: 
So the general solution is: 
Let us 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 example for finishing the assignment.
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 is 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.
It looks like a household waste dump, 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?
is the right side of the original equation, in this case:
The coefficient is the coefficient of the second derivative:
In practice, almost always, and our example is no exception.
Everything is clear, 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.
Let's find the main determinant of the system: 
If you have forgotten 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: this means that the system has a unique solution.
Finding the derivative: 
But that's not all, so far we have 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 remember in what form we were looking for a general solution to the inhomogeneous equation? In such:
The functions you need have just been found! 
All that remains is to perform the substitution and write down the answer:
Answer: common decision:
In principle, the answer could have expanded the parentheses.
A complete check of the answer is carried out according to the standard scheme, which was discussed in the lesson. Inhomogeneous 2nd order DEs. But the verification will not be easy, since it is necessary to find rather heavy derivatives and carry out cumbersome substitution. This is an unpleasant feature when you solve such diffusers.
Example 5
Solve a differential equation by varying arbitrary constants
This is an example for you to solve on your own. In fact, on the right side there is also a fraction. Let us remember the trigonometric formula; by the way, it will need to be applied during the solution.
The method of variation of arbitrary constants is the most universal method. It can solve any equation that can be solved method of selecting a particular solution based on the form of the right side. The question arises: why not use the method of variation of arbitrary constants there too? The answer is obvious: the selection of a particular solution, which was discussed in class Inhomogeneous second order equations, significantly speeds up the solution and shortens the recording - no fuss with determinants and integrals.
Let's look at two examples with Cauchy problem.
Example 6
Find a particular solution to the differential equation corresponding to the given initial conditions
, 
Solution: Again the fraction and exponent are in an interesting place.
We use the method of variation of arbitrary constants.
We'll find common decision appropriate homogeneous equations: 
– different real roots are obtained, so the general solution is:
General solution of inhomogeneous we look for equations in the form: , where – for now unknown functions.
Let's create a system:
In this case:
,
Finding derivatives: 
, 
Thus: 
Let's solve the system using Cramer's formulas:
, which means the system has a unique solution.
We restore the function by integration:
Used here method of subsuming a function under the differential sign.
We restore the second function by integration: 
This integral is solved variable replacement method:
From the replacement itself we express:
Thus: 
This integral can be found complete square extraction method, but in examples with diffusers I prefer to expand the fraction method of undetermined coefficients:
Both functions found:
As a result, the general solution to the inhomogeneous equation is:
Let's 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 differential equations of the second order.
Hold on, now we will find the derivative of the found general solution:
This is such a disgrace. It is not necessary to simplify it; it is easier to immediately create a system of equations. According to the initial conditions :
Let's substitute the found values of the constants 
to the general solution:
In the answer, the logarithms can be packed a little.
Answer: private solution: 
As you can see, difficulties may arise in integrals and derivatives, but not in the algorithm itself for the method of variation of arbitrary constants. It’s not me who intimidated you, it’s all Kuznetsov’s collection!
For relaxation, a final, simpler example for solving it yourself:
Example 7
Solve the Cauchy problem
, 
The example is simple, but creative, when you create a system, look at it carefully before deciding ;-),
As a result, the general solution is:
Let us find a particular solution corresponding to the initial conditions .
Let us 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.
Social politics The Bolsheviks were largely dictated by their class approach. By decree of November 10, 1917, the class system was destroyed, pre-revolutionary ranks, titles and awards were abolished. The election of judges has been established; secularization of civil states was carried out. Free education and medical care were established (decree of October 31, 1918). Women were given equal rights with men (decrees of December 16 and 18, 1917). The Decree on Marriage introduced the institution of civil marriage.
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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. It is necessary to prove that the amount
is a general solution to equation (1). Let us first prove that function (3) is a solution to 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 proven.
Let us prove the second statement: expression (3) is general solution of equation (1). We must prove that the arbitrary constants included in this expression can be selected so that the initial conditions are satisfied:
whatever the numbers are 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 can be represented in the form . Then, based on conditions (5), we will have
Let us solve this system and determine C 1 And C 2. Let's rewrite the system in the form:
Note that the determinant of this system is the Wronski determinant for the functions at 1 And at 2 at the point x=x 0. Since these functions are linearly independent by condition, the Wronski determinant is not equal to zero; therefore system (6) has a definite solution C 1 And C 2, i.e. there are such meanings C 1 And C 2, under which formula (3) determines the solution to equation (1) satisfying the given initial conditions. Q.E.D.
Let us move on to the general method of finding partial solutions to an inhomogeneous equation.
Let us write the general solution of the homogeneous equation (2)
We will look for a particular solution to the inhomogeneous equation (1) in the form (7), considering C 1 And C 2 like some as yet unknown functions from X.
Let us differentiate equality (7):
Let's select the functions you are looking for C 1 And C 2 so that the equality holds
If we take into account this additional condition, then the first derivative will take the form
Differentiating now this expression, we find:
Substituting into equation (1), we get
The expressions in the first two brackets become zero, since y 1 And y 2– 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 C 1 And C 2 satisfy equations (8) and (9). Let's create a system of equations from equations (8) and (9).
Since the determinant of this system is the Wronski determinant for linearly independent solutions y 1 And y 2 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 , from where, as a result of integration, we obtain . Next, we substitute the found functions into the formula, we obtain a general solution to the inhomogeneous equation, where are arbitrary constants.
A method for solving linear inhomogeneous differential equations of higher orders with constant coefficients by the method of variation of Lagrange constants is considered. The Lagrange method is also applicable to solving any linear inhomogeneous equations if the fundamental system of solutions to the homogeneous equation is known.
ContentSee also:
Lagrange method (variation of constants)
Consider a linear inhomogeneous differential equation with constant coefficients of arbitrary nth order:
(1)
.
The method of variation of a constant, which we considered for a first-order equation, is also applicable for higher-order equations.
The solution is carried out in two stages. In the first step, we discard the right-hand side and solve the homogeneous equation. As a result, we obtain a solution containing n arbitrary constants. At the second stage we vary the constants. That is, we believe that these constants are functions of the independent variable x and find the form of these functions.
Although we are considering equations with constant coefficients here, but Lagrange's method is also applicable to solving any linear inhomogeneous equations. To do this, however, the fundamental system of solutions to the homogeneous equation must be known.
Step 1. Solving the homogeneous equation
As in the case of first-order equations, we first look for the general solution of the homogeneous equation, equating the right-hand inhomogeneous side to zero:
(2)
.
The general solution to this equation is:
(3)
.
Here are arbitrary constants; - n linearly independent solutions of homogeneous equation (2), which form a fundamental system of solutions to this equation.
Step 2. Variation of constants - replacing constants with functions
In the second stage we will deal with the variation of constants. In other words, we will replace the constants with functions of the independent variable x:
.
That is, we are looking for a solution to the original equation (1) in the following form:
(4)
.
If we substitute (4) into (1), we get one differential equation for n functions.
In this case, we can connect these functions with additional equations. Then you get n equations from which n functions can be determined.
.
Additional equations can be written in various ways. But we will do this so that the solution has the simplest form. To do this, when differentiating, you need to equate to zero the terms containing derivatives of the functions.
.
Let's demonstrate this.
(5.1)
.
To substitute the proposed solution (4) into the original equation (1), we need to find the derivatives of the first n orders of the function written in the form (4). We differentiate (4) using the rules of differentiation of sum and product:
(6.1)
.
Let's group the members. First, we write down the terms with derivatives of , and then the terms with derivatives of :
.
Let's impose the first condition on the functions:
(5.2)
.
Then the expression for the first derivative with respect to will have a simpler form:
(6.2)
.
Using the same method, we find the second derivative:
Let's impose a second condition on the functions:
Then ,
And so on. In additional conditions, we equate terms containing derivatives of functions to zero.
Thus, if we choose the following additional equations for the functions: .
(5.k)
then the first derivatives with respect to will have the simplest form:
(6.k)
.
Here .
(1)
;
.
Find the nth derivative:
.
(6.n)
(7)
.
Substitute into the original equation (1):
(5.1)
;
(5.2)
;
(5.3)
;
. . . . . . .
Let us take into account that all functions satisfy equation (2): ;
Then the sum of terms containing zero gives zero. As a result we get: .
As a result, we received a system of linear equations for derivatives:
.
(5.n-1)
(7′) Lagrange's method is applicable to solve any linear inhomogeneous equations, if the fundamental system of solutions to the homogeneous equation (2) is known.
Examples
Solve equations using the method of variation of constants (Lagrange).
Solution of examples > > >
Solving higher order equations using the Bernoulli method
Solving linear inhomogeneous differential equations of higher orders with constant coefficients by linear substitution
The method of variation of an arbitrary constant, or the Lagrange method, is another way to solve first-order linear differential equations and the Bernoulli equation.
Linear differential equations of the first order are equations of the form y’+p(x)y=q(x). If there is a zero on the right side: y’+p(x)y=0, then this is a linear homogeneous 1st order equation. Accordingly, an equation with a non-zero right-hand side, y’+p(x)y=q(x), is heterogeneous 1st order linear equation.
Method of variation of an arbitrary constant (Lagrange method) is as follows:
1) We are looking for a general solution to the homogeneous equation y’+p(x)y=0: y=y*.
2) In the general solution, we consider C not a constant, but a function of x: C = C (x). We find the derivative of the general solution (y*)’ and substitute the resulting expression for y* and (y*)’ into the initial condition. From the resulting equation we find the function C(x).
3) In the general solution of the homogeneous equation, instead of C, we substitute the found expression C(x).
Let's look at examples of the method of varying an arbitrary constant. Let's take the same tasks as in, compare the progress of the solution and make sure that the answers obtained coincide.
1) y’=3x-y/x
Let's rewrite the equation in standard form (unlike Bernoulli's method, where we needed the notation form only to see that the equation is linear).
y’+y/x=3x (I). Now we proceed according to plan.
1) Solve the homogeneous equation y’+y/x=0. This is an equation with separable variables. Imagine y’=dy/dx, substitute: dy/dx+y/x=0, dy/dx=-y/x. We multiply both sides of the equation by dx and divide by xy≠0: dy/y=-dx/x. Let's integrate:
2) In the resulting general solution of the homogeneous equation, we will consider C not a constant, but a function of x: C=C(x). From here
We substitute the resulting expressions into condition (I):
Let's integrate both sides of the equation:
here C is already some new constant.
3) In the general solution of the homogeneous equation y=C/x, where we assumed C=C(x), that is, y=C(x)/x, instead of C(x) we substitute the found expression x³+C: y=(x³ +C)/x or y=x²+C/x. We got the same answer as when solving by Bernoulli's method.
Answer: y=x²+C/x.
2) y’+y=cosx.
Here the equation is already written in standard form; there is no need to transform it.
1) Solve the homogeneous linear equation y’+y=0: dy/dx=-y; dy/y=-dx. Let's integrate:
To obtain a more convenient form of notation, we take the exponent to the power of C as the new C:
This transformation was performed to make it more convenient to find the derivative.
2) In the resulting general solution of the linear homogeneous equation, we consider C not a constant, but a function of x: C=C(x). Under this condition
We substitute the resulting expressions y and y’ into the condition:
Multiply both sides of the equation by
We integrate both sides of the equation using the integration by parts formula, we get:
Here C is no longer a function, but an ordinary constant.
3) In the general solution of the homogeneous equation
substitute the found function C(x):
We got the same answer as when solving by Bernoulli's method.
The method of variation of an arbitrary constant is also applicable to solve.
y'x+y=-xy².
We bring the equation to standard form: y’+y/x=-y² (II).
1) Solve the homogeneous equation y’+y/x=0. dy/dx=-y/x. We multiply both sides of the equation by dx and divide by y: dy/y=-dx/x. Now let's integrate:
We substitute the resulting expressions into condition (II):
Let's simplify:
We obtained an equation with separable variables for C and x:
Here C is already an ordinary constant. During the integration process, we wrote simply C instead of C(x), so as not to overload the notation. And at the end we returned to C(x), so as not to confuse C(x) with the new C.
3) In the general solution of the homogeneous equation y=C(x)/x we substitute the found function C(x):
We got the same answer as when solving it using the Bernoulli method.
Self-test examples:
1. Let's rewrite the equation in standard form: y’-2y=x.
1) Solve the homogeneous equation y’-2y=0. y’=dy/dx, hence dy/dx=2y, multiply both sides of the equation by dx, divide by y and integrate:
From here we find y:
We substitute the expressions for y and y’ into the condition (for brevity we will use C instead of C(x) and C’ instead of C"(x)):
To find the integral on the right side, we use the integration by parts formula:
Now we substitute u, du and v into the formula:
Here C =const.
3) Now we substitute homogeneous into the solution