1000 Solved Problems in Modern Physics

1.3 Solutions 63

Differentiating (1) twice

d^4 y

dx^4

+

d^3 y

dx^3

− 2

d^2 y

dx^2

=8 cosh 2x (3)

Multiply (1) by (4) and subtract the resulting equation from (3)

d^4 y

dx^4

+

d^3 y

dx^3

−

6d^2 y

dx^2

−

4dy

dx

+ 8 y=0(4)

D^4 +D^3 − 6 D^2 − 4 D+ 8 = 0

(D−1)(D−2)(D+2)^2 = 0

D= 1 , 2 ,− 2 ,− 2

The complete solution of (4) is

y=C 1 ex+C 3 e^2 x+C 2 e−^2 x+C 4 xe−^2 x

=U+C 3 e^2 x+C 4 xe−^2 x

=U+V

V=C 3 e^2 x+C 4 xe−^2 x (5)

Inserting (5) in (1), writing 2 cosh 2 x =e^2 x+e−^2 xand comparing the

coefficients ofe^2 xande−^2 x, we findC 3 =^14 andC 4 =−^13. Thus the complete

solution of (1) is

y=C 1 ex+C 2 e−^2 x+

1

4

e^2 x−

1

3

xe−^2 x

1.66

xdy

dx

−y=x^2

dy

dx

−

y

x

=x

The standard equation is

dy

dx

+Py=Q

∴P=−

1

x

;Q=x

yexp

(∫

pdx

)

=

[∫

Qexp

(∫

pdx

)]

dx+C

yexp

(

−

∫

1

x

dx

)

=

[∫

xexp

(

−

∫

1

x

dx

)]

+C

yexp(−lnx)=

[∫

xexp(−lnx)

]

+C

yx−^1 =

∫

xx−^1 dx+C

y=x^2 +Cx