Higher Engineering Mathematics, Sixth Edition

486 Higher Engineering Mathematics

Now try the following exercise

Exercise 189 Further problems on

differential equations of the form

a

d^2 y

dx^2

+b

dy

dx

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

constant or polynomial.

In Problems 1 and 2, find the general solutions of

the given differential equations.

1. 2

d^2 y

dx^2

+ 5

dy

dx

− 3 y= 6

[

y=Ae

1

2 x+Be−^3 x− 2

]

2. 6

d^2 y

dx^2

+ 4

dy

dx

− 2 y= 3 x− 2

[

y=Ae

1

3 x+Be−x− 2 −^32 x

]

In Problems 3 and 4 find the particular solutions of

the given differential equations.

3. 3

d^2 y

dx^2

+

dy

dx

− 4 y=8; whenx=0,y=0and

dy

dx

=0.

[

y=^27 (3e−

4

3 x+4ex)− 2

]

4. 9

d^2 y

dx^2

− 12

dy

dx

+ 4 y= 3 x−1; when x=0,

y=0and

dy

dx

=−

4

3

[

y=−

(

2 +^34 x

)

e

2

3 x+ 2 +^34 x

]

5. The chargeqin an electric circuit at timetsat-

isfies the equationL

d^2 q

dt^2

+R

dq

dt

+

1

C

q=E,

whereL,R,CandEare constants. Solve the

equation givenL= 2 H,C= 200 × 10 −^6 Fand

E=250V, when (a)R= 200 and (b)Ris

negligible. Assume that whent=0,q=0and

dq

dt

= 0

⎡

⎢

⎢

⎣

(a) q=

1

20

−

(

5

2

t+

1

20

)

e−^50 t

(b) q=

1

20

( 1 −cos50t)

⎤

⎥

⎥

⎦

6. In a galvanometer the deflectionθ satisfies
   the differential equation

d^2 θ

dt^2

+ 4

dθ

dt

+ 4 θ=8.

Solve the equation forθgiven that whent=0,

θ=

dθ

dt

=2. [θ= 2 (te−^2 t+ 1 )]

51.4 Workedproblemsondifferential

equations of the form

a

d^2 y

dx^2

+b

dy

dx

+cy=f(x)where

f(x)is an exponential function

Problem 4. Solve the equation

d^2 y

dx^2

− 2

dy

dx

+y=3e^4 xgiven the boundary

conditions that whenx=0,y=−^23 and

dy

dx

= (^413)
Using the procedure of Section 51.2:
(i)
d^2 y
dx^2
− 2
dy
dx
+y=3e^4 x in D-operator form is
(D^2 −2D+ 1 )y=3e^4 x.
(ii) Substituting m for D gives the auxiliary
equation m^2 − 2 m+ 1 =0. Factorizing gives:
(m− 1 )(m− 1 )=0, from which,m=1 twice.
(iii) Since the roots are real and equal the C.F.,
u=(Ax+B)ex.
(iv) Let the particular integral, v=ke^4 x (see
Table 51.1(c)).
(v) Substitutingv=ke^4 xinto
(D^2 −2D+ 1 )v=3e^4 xgives:
(D^2 −2D+ 1 )ke^4 x=3e^4 x
i.e. D^2 (ke^4 x)−2D(ke^4 x)+ 1 (ke^4 x)=3e^4 x
i.e. 16 ke^4 x− 8 ke^4 x+ke^4 x=3e^4 x
Hence 9ke^4 x=3e^4 x, from which,k=^13
Hence the P.I.,v=ke^4 x=^13 e^4 x.
(vi) The general solution is given byy=u+v,i.e.
y=(Ax+B)ex+^13 e^4 x.
(vii) Whenx=0,y=−^23 thus