differential equation

458 DIFFERENTIAL EQUATIONS
When x =−1,
−6 =−3A, from which, A = 2
When x = 2,
3 = 3B, from which, B = 1
∫
3x − 3
Hence
(x + 1)(x − 2) dx
∫ [ 2
=
x + 1 + 1 ]
dx
x − 2
(iii) Integrating factor
= 2ln(x + 1) + ln(x − 2)
= ln[(x + 1) 2 (x − 2)]
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 1
(x − 2)
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
(x + 1) 3
i.e. y =
3(x + 1) 2 (x − 2)
and hence the 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. cot x dy
dx = 1 − 2y,giveny = 1 when x = π 4 .
[y = 1 2 + cos2 x]
2. t dθ + sec t(t sin t + cos t)θ = sec t, given
dt
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 ln(x + 2),
x
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
+ ky = a sin bx
dx
is given by:
(
y =
given y = 1 when x = 0.
)
a
k 2 + b 2 (k sin bx − b cos bx)
( k 2 + b 2 )
+ ab
+
k 2 + b 2 e −kx ,
6. The equation dv =−(av + bt), where a and
dt
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
7. In an alternating current circuit containing
resistance R and inductance L the current i is
given by: Ri + L di
dt = E 0 sin ωt. Giveni = 0
when t = 0, show that the solution of the
equation is given by:
(
i =
)
E 0
R 2 + ω 2 L 2 (R sin ωt − ωL cos ωt)
( )
E0 ωL
+
R 2 + ω 2 L 2 e −Rt/L