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CHAPTER 4. HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS 4.5. The Method of Variation of Parameters

In this section we consider a general method of determining a particular solution of the equation

a 0 (x) d 2 y

dx 2 + a 1 (x) dy

dx + a 2 (x)y = f (x); (1) where the functions a 0 (x); a 1 (x) and a 2 (x) are continuous on the interval [a; b].

Suppose that y 1 and y 2 are linearly independent solutions of the corresponding homogeneous equation

a 0 (x) d 2 y

dx 2 + a 1 (x) dy

dx + a 2 (x)y = 0: (2)

Then the complementary function of equation (2) is y c = c 1 y 1 (x) + c 2 y 2 (x)

The procedure in the method of variation of parameters is to replace the ar- bitrary constants c 1 and c 2 by the functions v 1 (x) and v 2 (x) which will be determined so that the resulting function

y p = v 1 (x)y 1 (x) + v 2 (x)y 2 (x) (3) will be a particular solution of equation (1): It can be seen that to determine v 1 (x) and v 2 (x); we have the following system of equations

v 1 0 (x)y 1 (x) + v 0 2 (x)y 2 (x) = 0 (4) v 1 0 (x)y 0 1 (x) + v 0 2 (x)y 2 0 (x) = f (x)

a 0 (x) for unknown functions v 1 0 and v 2 0 :

Since y 1 and y 2 are linearly independent solutions of equation (2); the determi- nant of coe¢ cients of this system

W (y 1 (x); y 1 (x)) = y 1 (x) y 2 (x) y 1 0 (x) y 2 0 (x)

= y 1 y 0 2 y 2 y 1 0 6= 0:

By Cramer’s Rule, the solution of system (4) is obtained as

v 0 1 (x) = 1 W (y 1 (x); y 1 (x))

0 y 2 (x)

f (x)

a

0

(x) y 0 2 (x)

1

(2)

v 0 2 (x) = 1 W (y 1 (x); y 1 (x))

y 1 (x) 0 y 0 1 (x) a f (x)

0

(x)

Integrating v 0 1 (x) and v 0 2 (x); we can …nd v 1 (x) and v 2 (x): So, the particular solution of equation (1) is

y p (x) = v 1 (x)y 1 (x) + v 2 (x)y 2 (x):

Remark 1. This method can be apply higher order di¤erential equations. Let us consider

d 3 y

dx 3 + a(x) d 2 y

dx 2 + b(x) dy

dx + c(x)y = f (x) The particular solution of this equation will be in the form

y p = v 1 (x)y 1 (x) + v 2 (x)y 2 (x) + v 3 (x)y 3 (x);

where y 1 ; y 2 and y 3 are linearly independent solutions of corresponding homo- geneous equation. Similarly, from the following system v 1 0 ; v 0 2 and v 3 0 can be found:

v 1 0 y 1 + v 0 2 y 2 + v 0 3 y 3 = 0 v 1 0 y 1 0 + v 0 2 y 0 2 + v 0 3 y 0 3 = 0 v 0 1 y 1 00 + v 0 2 y 00 2 + v 3 0 y 3 00 = f (x)

Example 1. Find the general solution of the di¤erential equation d 2 y

dx 2 + y = cos ecx

Solution. It can be easily seen that the complementary function is y c = c 1 cos x + c 2 sin x

So, the particular solution is in the form

y p = v 1 (x) cos x + v 2 (x) sin x:

v 1 and v 2 should be satisfy the following system:

v 0 1 cos x + v 0 2 sin x = 0 v 1 0 ( sin x) + v 0 2 (cos x) = 1

sin x The determinanf of the coe¢ cients of this system is

cos x sin x

sin x cos x = 1 6= 0:

So,

v 1 0 (x) = 0 sin x

1

sin x cos x = 1

2

(3)

and

v 2 0 (x) = cos x 0

sin x sin x 1 = cos x sin x Integrating these equations we obtain

v 1 (x) = x and

v 2 (x) = ln(sin x):

Thus the particular solution is found as

y p = x cos x + ln(sin x) sin x The general solution of given di¤erential equation is

y = c 1 cos(x) + c 2 sin x x cos x + ln(sin x) sin x:

Example. Find the general solutions of following di¤erential equations.

1)

d 2 y dx 2 4 dy

dx + 3y = 1 1 + e x 2)

d 3 y dx 3 + d 2 y

dx 2 dy

dx y = 1 cos x

3

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