Page 76 - Advanced engineering mathematics

56 CHAPTER 2 Linear Second-Order Equations

Once we have u 1 and u 2 , we have a particular solution Y p = u 1 y 1 + u 2 y 2 , and the general
solution of y + p(x)y + q(x)y = f (x) is

y = c 1 y 1 + c 2 y 2 + Y p .

EXAMPLE 2.10
Find the general solution of
y + 4y = sec(x)

for −π/4 < x <π/4.
2
The characteristic equation of y + 4y = 0is λ + 4 = 0 with complex roots λ =±2i.Two

linearly independent solutions of the associated homogeneous equation y + 4y = 0are

y 1 (x) = cos(2x) and y 2 (x) = sin(2x).
Now look for a particular solution of the nonhomogeneous equation. First compute the
Wronskian

cos(2x) sin(2x)
2 2
W(x) = = 2(cos (x) + sin (x)) = 2.
−2sin(2x) 2cos(2x)

Use equations (2.11) with f (x) = sec(x) to obtain

sin(2x)sec(x)
u 1 (x) =− dx
2

2sin(x)cos(x)sec(x)
=− dx
2

sin(x)cos(x)
=− dx
cos(x)

=− sin(x)dx = cos(x)
and
cos(2x)sec(x)

u 2 (x) = dx
2
(2cos (x) − 1)
2
= dx
2cos(x)
1

= cos(x) − sec(x) dx
2
1
= sin(x) − ln|sec(x) + tan(x)|.
2
This gives us the particular solution
Y p (x) = u 1 (x)y 1 (x) + u 2 (x)y 2 (x)

1
= cos(x)cos(2x) + sin(x) − ln|sec(x) + tan(x)| sin(2x).
2
The general solution of y + 4y = sec(x) is

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

1
+ cos(x)cos(2x) + sin(x) − ln|sec(x) + tan(x)| sin(2x).
2

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October 14, 2010 14:12 THM/NEIL Page-56 27410_02_ch02_p43-76