492 Higher Engineering Mathematics
Equating coefficients of e
x sin 2x gives:
−3C − 4D − 2C + 4D + 2C = 0
i.e. −3C = 0, from which, C = 0.
Equating coefficients of e x cos 2x gives:
4C − 3D − 4C − 2D + 2D = 3
i.e. −3D = 3, from which, D =−1.
Hence the P.I., υ = e x (−cos 2x).
(vi) The general solution, y = u + v, i.e.
y = e
x
(A cos x +B sinx) − e
x cos 2x
(vii) When x = 0, y = 2 thus
2 = e 0 (A cos 0 + B sin 0)
− e 0 cos 0
i.e.
2 = A − 1, from which, A = 3
d y
dx
= e x (− A sin x + B cos x)
+ e x (A cos x + B sin x)
− [e x (−2 sin 2x) + e x cos 2x]
When
x = 0,
d y
dx
= 3
thus
3 = e 0 (− A sin 0 + B cos 0)
+ e 0 (A cos 0 + B sin 0)
− e 0 (−2 sin0) − e 0 cos 0
i.e.
3 = B + A − 1, from which,
B = 1, since A = 3
Hence the particular solution is
y = e
x (3 cos x + sin x) − e
x cos 2x
Now try the following exercise
Exercise 192 Further problems on second
order differential equations of the form
a
d 2 y
dx 2 + b
dy
dx
+ cy = f (x) where f (x) is a sum or
product
In Problems 1 to 4, find the general solutions of the
given differential equations.
1. 8
d 2 y
dx 2 − 6
d y
dx
+ y = 2x + 40 sin x
⎡
⎣
y = Ae
x
4 + Be
x
2 + 2x + 12
+
8
17
(6 cos x − 7 sin x)
⎤
⎦
2.
d 2 y
dθ 2 − 3
d y
dθ
+ 2y = 2 sin 2 θ − 4 cos 2 θ
y = Ae 2θ + Be θ +
1
2 (sin 2 θ + cos 2 θ)
3.
d 2 y
dx 2 +
d y
dx
− 2y = x 2 + e 2x
y = Ae x + Be −2x −
3
4
−
1
2 x −
1
2 x 2 +
1
4 e 2x
4.
d 2 y
dt 2 − 2
d y
dt
+ 2y = e t sin t
y = e t (A cos t + B sin t ) −
t
2 e t cos t
In Problems 5 to 6 find the particular solutions of
the given differential equations.
5.
d 2 y
dx 2 − 7
d y
dx
+ 10y = e 2x + 20; when x = 0,
y = 0 and
d y
dx
=−
1
3
y =
4
3
e 5x −
10
3
e 2x −
1
3
xe 2x + 2
6. 2
d 2 y
dx 2 −
d y
dx
− 6y = 6e x cos x; when x = 0,
y =−
21
29
and
d y
dx
=−6
20
29
⎡
⎣
y = 2e
−
3
2 x
− 2e 2x
+
3e x
29
(3 sin x − 7 cos x)
⎤
⎦