1000 Solved Problems in Modern Physics

1.3 Solutions 65

d^4 y

dx^4

+

4d^2 y

dx^2

=−8 cos(2x)(2)

Multiply (1) by 4 and add to (2),

d^4 y

dx^4

+

8d^2 y

dx^2

+ 16 y= 0

D^4 + 8 D^2 + 16 = 0

(D^2 +4)^2 = 0

D=± 2 i

y=C 1 sin 2x+C 3 xsin 2x+C 2 cos 2x+C 4 xcos 2x

=U+C 3 xsin 2x+C 4 xcos 2x

=U+V

Y=V=C 3 xsin 2x+C 4 xcos 2x (3)

Use (3) in (1) and compare the coefficients of sin 2xand cos 2xto find

C 3 = 1 /2 andC 4 =0. Thus the complete solution is

y=C 1 sin 2x+C 2 cos 2x+

1

2

xsin 2x

1.69

dy

dx

+

3 y

x+ 2

=x+ 2

This equation is of the form

dy

dx

+yp(x)=Q(x)

withP=x+^32 andQ=x+ 2

The solution is obtained from

yexp

(∫

p(x)dx=

[∫

Q(x)exp

(∫

p(x)dx

)]

dx+C

Now

∫

p(x)dx= 3

∫

dx

x+ 2

=3ln(x+2)=ln(x+2)^3

∴yexp (ln(x+2)^3 )=

[∫

(x+2) exp (ln (x+2)^3 )

]

dx+C

y(x+2)^3 =

∫

(x+2)^4 dx+C

=

(x+2)^5

5

+C

ory=

(x+2)^2

5

+C

y=2 whenx=− 1

ThereforeC=^95

The complete solution is

y=

(x+2)^2

5

+

9

5

=

x^2 + 4 x+ 13

5