Linear first order differential equations 459
Let
3x − 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)
When x =−1,
−6 =−3 A, from which, A = 2
When x = 2,
3 = 3B, from which, B = 1
Hence
3x − 3
(x + 1)(x − 2)
dx
=
2
x + 1
+
1
x − 2
dx
= 2 ln(x + 1) + ln (x − 2)
= ln [(x + 1)
2
(x − 2)]
(iii) Integrating factor
e
P dx
= e
ln[(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) =
1
3 (x + 1)
3
+ c
(b) When x =−1, y = 5 thus 5(0)(−3) = 0 + c, from
which, c = 0.
Hence y(x + 1) 2 (x − 2) =
1
3 (x + 1) 3
i.e. y =
(x + 1) 3
3(x + 1) 2 (x − 2)
and hence the particular solution is
y =
(x + 1)
3(x − 2)
Now try the following exercise
Exercise 183 Further problems on linear
first order differential equations
In problems 1 and 2, solve the differential equations.
1. cot x
dy
dx
= 1 − 2y, given y = 1 when x =
π
4
.
[y =
1
2 + cos 2 x]
2. t
dθ
dt
+ sec t (t sin t + cos t )θ = sec t , given
t = π when θ = 1.
θ =
1
t
(sin t − π cos t )
3. Given the equation x
dy
dx
=
2
x + 2
− y show
that the particular solution is y =
2
x
ln (x + 2),
given the boundary conditions that x =−1
when y = 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
when y = 0.
5. Show that the solution of the differential
equation
dy
dx
+ ky = a sin bx
is given by:
y =
a
k 2 + b 2
(k sin bx − b cos bx)
+
k 2 + b 2 + ab
k 2 + b 2
e
−kx
,
given y = 1 when x = 0.
6. The equation
dv
dt
=−(av + bt ), where a and
b are constants, represents an equation of
motion when a particle moves in a resisting
medium. Solve the equation for v given that
v = u when t = 0.
v =
b
a 2 −
bt
a
+
u −
b
a 2
e
−at