Linear first order differential equations 459

Let

3 x− 3

(x+ 1 )(x− 2 )

≡

A

(x+ 1 )

+

B

(x− 2 )

≡

A(x− 2 )+B(x+ 1 )

(x+ 1 )(x− 2 )

from which, 3x− 3 =A(x− 2 )+B(x+ 1 )

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 183 Further problemson linear
first order differential equations
In problems 1 and 2, solve the differential equa-
tions.

1. cotx

dy

dx

= 1 − 2 y,giveny=1whenx=

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 equation x

dy

dx

=

2

x+ 2

−y show

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 that x= 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=1whenx=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

]