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.
- 2
d^2 y
dx^2
+ 5
dy
dx
− 3 y= 6
[
y=Ae
1
2 x+Be−^3 x− 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
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
]
- 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
]
- 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)
⎤
⎥
⎥
⎦
- 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