Higher Engineering Mathematics, Sixth Edition

492 Higher Engineering Mathematics

Equating coefficients of exsin2xgives:

− 3 C− 4 D− 2 C+ 4 D+ 2 C= 0

i.e.− 3 C=0, from which,C=0.

Equating coefficients of excos2xgives:

4 C− 3 D− 4 C− 2 D+ 2 D= 3

i.e.− 3 D=3, from which,D=−1.

Hence the P.I.,υ=ex(−cos2x).

(vi) The general solution,y=u+v,i.e.

y=ex(Acosx+Bsinx)−excos2x

(vii) Whenx=0,y=2 thus

2 =e^0 (Acos 0+Bsin0)

−e^0 cos0

i.e. 2 =A− 1 ,from which,A= 3

dy

dx

=ex(−Asinx+Bcosx)

+ex(Acosx+Bsinx)

−[ex(−2sin2x)+excos2x]

When x= 0 ,

dy

dx

= 3

thus 3 =e^0 (−Asin 0+Bcos0)

+e^0 (Acos 0+Bsin0)

−e^0 (−2sin0)−e^0 cos0

i.e. 3 =B+A− 1 ,from which,

B= 1 ,since A= 3

Hence the particular solution is

y=ex(3 cosx+sinx)−excos2x

Now try the following exercise

Exercise 192 Further problems on second

order differential equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)wheref(x)is a sum or

product

In Problems 1 to 4, find the general solutionsof the

given differential equations.

1. 8

d^2 y

dx^2

− 6

dy

dx

+y= 2 x+40sinx

⎡

⎣

y=Ae

x

(^4) +Be
x
(^2) + 2 x+ 12

• 8
  17
  (6cosx−7sinx)
  ⎤
  ⎦

2. d^2 y
   dθ^2
   − 3
   dy
   dθ

• 2 y=2sin2θ−4cos2θ
  [
  y=Ae^2 θ+Beθ+^12 (sin2θ+cos2θ)
  ]

3. d^2 y
   dx^2

• dy
  dx
  − 2 y=x^2 +e^2 x
  [
  y=Aex+Be−^2 x−^34
  −^12 x−^12 x^2 +^14 e^2 x
  ]

4. d^2 y
   dt^2
   − 2
   dy
   dt

• 2 y=etsint
  [
  y=et(Acost+Bsint)− 2 tetcost
  ]
  In Problems 5 to 6 find the particular solutions of
  the given differential equations.

5. d^2 y
   dx^2
   − 7
   dy
   dx

• 10 y=e^2 x+20; when x=0,
  y=0and
  dy
  dx
  =−
  1
  3
  [
  y=
  4
  3
  e^5 x−
  10
  3
  e^2 x−
  1
  3
  xe^2 x+ 2
  ]

6. 2

d^2 y

dx^2

−

dy

dx

− 6 y=6excosx;whenx=0,

y=−

21

29

and

dy

dx

=− 6

20

29

⎡

⎣

y=2e−

(^32) x
−2e^2 x

• 3ex
  29
  (3sinx−7cosx)
  ⎤
  ⎦