1000 Solved Problems in Modern Physics

66 1 Mathematical Physics

1.70 (i)d

(^2) y
dx^2

−

4dy

dx

+ 4 y= 8 x^2 − 4 x−4(1)

Replace the RHS member by zero to get the auxiliary solution.

D^2 − 4 D+ 4 = 0

The roots areD=2 and 2. Therefore the auxiliary solution is

y=Ae^2 x+Bxe^2 x (2)

Complete solution is

y=(A+Bx)e^2 x+Cx^2 +Dx+E (3)

The derivatives are

dy

dx

=(2A+ 2 Bx+B)e^2 x+ 2 Cx+D (4)

d^2 y/dx^2 =4(A+B+Bx)e^2 x+ 2 C (5)

Use (3), (4) and (5) in (1) and compare the coefficients of like terms. We

get three equations. Two more equations are obtained from the conditions

y=−2 andddxy=0 whenx=0.

Solving the five equations we get,A=−3,B=3,C=2,D=3 and

E=1. Hence the complete solution isy=3(x−1)e^2 x+ 2 x^2 + 3 x+ 1

(ii) d

(^2) y
dx^2

• 4 y=sinx (1)
  Replace the RHS member by zero and write down the auxiliary equation
  D^2 + 4 = 0
  The roots are± 2 i. The auxiliary solution is
  Y=Acos 2x+Bsin 2x
  The complete solution is
  Y=Acos 2x+Bsin 2x+Csinx (2)
  d^2 y
  dx^2
  =−4(Acos 2x+Bsin 2x)−Csinx (3)
  Substitute (2) and (3) in (1) to findC= 1 /3. Thus
  y=Acos 2x+bsin 2x+

1

3

sinx

1.71 y′′′−y′′+y′−y= 0
Auxiliary equation is
D^3 −D^2 +D− 1 = 0
(D−1)(D^2 +1)=0 The roots areD= 1 ,±i
Thesolutionis
y=Asinx+Bcosx+cex