1000 Solved Problems in Modern Physics

58 1 Mathematical Physics

1.58

d^2 y

dx^2

+m^2 y=cosbx (1)

Replacing the right-hand member by zero,

d^2 y

dx^2

+m^2 y= 0. (2)

Solving, we get the complimentary function

y=C 1 sinmx+C 2 cosmx=U. (3)

Differentiating (1) twice, we get

d^4 y

dx^4

+m^2

d^2 y

dx^2

=−b^2 cosbx. (4)

Multiply (1) byb^2 and adding the result to (4) gives

d^4 y

dx^4

+(m^2 +b^2 )

d^2 y

dx^2

+b^2 m^2 y= 0. (5)

The complete solution of (5) is

y=C 1 sinmx+C 2 cosmx+C 3 sinbx+C 4 cosbx

ory=U+C 3 sinbx+C 4 cosbx=U+V

We shall now determineC 3 andC 4 so thatC 3 sinbx+C 4 cosbxshall be a

particular solutionVof (1)

Substituting

y=C 3 sinbx+C 4 cosbx,

dy

dx

=C 3 bcosbx−C 4 bsinbx,

d^2 y

dx^2 =−C^3 b

(^2) sin bx−C 4 b (^2) cos bx in (1), we get
C 4 (m^2 −b^2 ) cosbx+C 3 (m^2 −b^2 )sinbx=cosbx
Equating the coefficients of like terms in this identity we get
C 4 (m^2 −b^2 )= 1 →C 4 =

1

m^2 −b^2

C 3 (m^2 −b^2 )= 0 →C 3 = 0

Hence a particular solution of (1) is

V=

cosbx

m^2 −b^2

and the complete solution is

y= 0 +V=C 1 sinmx+C 2 cosmx+

cosbx

m^2 −b^2

1.59
d^2 y
dx^2

− 5

dy

dx

+ 6 y=x (1)

Replace the right-hand member by zero to form the auxiliary equation

D^2 − 5 D+ 6 =0(2)

The roots areD=2 and 3. The solution is

y=C 1 e^2 x+C 2 e^3 x=0(3)