Higher Engineering Mathematics

484 DIFFERENTIAL EQUATIONS

4. 9

d^2 y

dx^2

− 12

dy

dx

+ 4 y= 3 x−1; whenx=0,

y=0 and

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 F

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

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

and

dq

dt

=0.

⎡

⎢

⎢

⎣

(a) q=

1

20

−

(

5

2

t+

1

20

)

e−^50 t

(b) q=

1

20

(1−cos 50t)

⎤

⎥

⎥

⎦

6. In a galvanometer the deflectionθsatisfies 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 Worked problems on differential

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 xin D-operator form is
(D^2 −2D+1)y=3e^4 x.
(ii) Substituting m for D gives the auxiliary
equationm^2 − 2 m+ 1 =0. Factorising 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
−^23 =(0+B)e^0 +^13 e^0 , from which,B=−1.
dy
dx
=(Ax+B)ex+ex(A)+^43 e^4 x.
Whenx=0,
dy
dx
= 4
1
3
, thus
13
3
=B+A+
4
3
from which,A=4, sinceB=−1.
Hence the particular solution is:
y=( 4 x− 1 )ex+^13 e^4 x
Problem 5. Solve the differential equation
2
d^2 y
dx^2
−
dy
dx
− 3 y=5e
3
2 x.
Using the procedure of Section 51.2:
(i) 2
d^2 y
dx^2
−
dy
dx
− 3 y=5e
3
2 xin D-operator form is
(2D^2 −D−3)y=5e
3
2 x.
(ii) Substituting m for D gives the auxiliary
equation 2m^2 −m− 3 =0. Factorising gives:
(2m−3)(m+1)=0, from which, m=^32 or
m=−1. Since the roots are real and different
then the C.F.,u=Ae
3
2 x+Be−x.
(iii) Since e
3
2 x appears in the C.F. and in the
right hand side of the differential equation, let