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).
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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 to 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.
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
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.
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:
.
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:
(5.1)
.
Then the expression for the first derivative with respect to will have a simpler form:
(6.1)
.
Using the same method, we find the second derivative:
.
Let's impose a second condition on the functions:
(5.2)
.
Then
(6.2)
.
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 .
Find the nth derivative:
(6.n)
.
Substitute into the original equation (1):
(1)
;
.
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:
(7)
.
As a result, we received a system of linear equations for derivatives:
(5.1)
;
(5.2)
;
(5.3)
;
. . . . . . .
(5.n-1) ;
(7′) .
Solving this system, we find expressions for derivatives as a function of x. Integrating, we get:
.
Here are constants that no longer depend on x. Substituting into (4), we obtain a general solution to the original equation.
Note that to determine the values of the derivatives, we never used the fact that the coefficients a i are constant. That's why 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
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 of replacing arbitrary constants c k in the general solution
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
for auxiliary functions c k (t) , whose derivatives satisfy the linear algebraic system
The determinant of system (1) is the Wronskian of the 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 integration constants, 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 to 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 to the corresponding homogeneous equation, written in the form of a matrix, and the vector function , which replaced the vector of arbitrary constants, is defined by the relation . The required particular solution (with zero initial values at t = t 0 looks like
For a system with constant coefficients, the last expression is simplified:
Matrix Z(t)Z− 1 (τ) called Cauchy matrix operator L = A(t) .