458 DIFFERENTIAL EQUATIONS

Whenx=−1,

− 6 =− 3 A, from which,A= 2

Whenx=2,

3 = 3 B, from which,B= 1

Hence

∫

3 x− 3

(x+1)(x−2)

dx

=

∫ [

2

x+ 1

+

1

x− 2

]

dx

=2ln(x+1)+ln(x−2)

=ln[(x+1)^2 (x−2)]

(iii) Integrating factor

e

∫

Pdx=eln [(x+1)^2 (x−2)]=(x+1) (^2) (x−2)
(iv) Substituting in equation (3) gives:
y(x+1)^2 (x−2)

∫
(x+1)^2 (x−2)
1
x− 2
dx

∫
(x+1)^2 dx
(v)Hence the general solution is:
y(x+1)^2 (x−2)=^13 (x+1)^3 +c
(b) Whenx=−1,y=5 thus 5(0)(−3)= 0 +c, from
which,c=0.
Hencey(x+1)^2 (x−2)=^13 (x+1)^3
i.e.y=
(x+1)^3
3(x+1)^2 (x−2)
and hencethe particular solution is
y=
(x+ 1 )
3 (x− 2 )
Now try the following exercise.
Exercise 184 Further problems on linear
first order differential equations
In problems 1 and 2, solve the differential
equations

1. cotx

dy

dx

= 1 − 2 y,giveny=1 whenx=

π

4

.

[y=^12 +cos^2 x]

2.t

dθ

dt

+sect(tsint+cost)θ=sect,given

t=πwhenθ=1.

[

θ=

1

t

(sint−πcost)

]

3. Given the equationx

dy

dx

=

2

x+ 2

−yshow

that the particular solution isy=

2

x

ln(x+2),

given the boundary conditions thatx=− 1

wheny=0.

4. Show that the solution of the differential
   equation
   dy
   dx

−2(x+1)^3 =

4

(x+1)

y

is y=(x+1)^4 ln(x+1)^2 , given thatx= 0

wheny=0.

5. Show that the solution of the differential
   equation
   dy
   dx

+ky=asinbx

is given by:

y=

(

a

k^2 +b^2

)

(ksinbx−bcosbx)

+

(

k^2 +b^2 +ab

k^2 +b^2

)

e−kx,

giveny=1 whenx=0.

6. The equation

dv

dt

=−(av+bt), whereaand

bare constants, represents an equation of

motion when a particle moves in a resisting

medium. Solve the equation forvgiven that

v=uwhent=0.

[

v=

b

a^2

−

bt

a

+

(

u−

b

a^2

)

e−at

]

7. In an alternating current circuit containing
   resistanceRand inductanceLthe currentiis
   given by:Ri+L

di

dt

=E 0 sinωt.Giveni= 0

whent=0, show that the solution of the

equation is given by:

i=

(

E 0

R^2 +ω^2 L^2

)

(Rsinωt−ωLcosωt)

+

(

E 0 ωL

R^2 +ω^2 L^2

)

e−Rt/L