Second order differential equations of the form a
d 2 y
dx 2 + b
dy
dx + cy = f (x) 491
(v) Substituting v into
(D 2 + D −6)v = 12x − 50 sin x gives:
(D
2
+ D − 6)(ax + b + c sin x + d cos x)
= 12x − 50 sin x
D(ax + b + c sin x + d cos x)
= a + c cos x − d sin x
D
2
(ax + b + c sin x + d cos x)
= −c sin x − d cos x
Hence (D 2 + D − 6)(v)
= (−c sin x − d cos x) + (a + c cos x
− d sin x) − 6(ax + b + c sin x + d cos x)
= 12x − 50 sin x
Equating constant terms gives:
a − 6b = 0
( 1 )
Equating coefficients of x gives: −6a = 12, from
which, a =−2.
Hence, from (1), b =−
1
3
Equating the coefficients of cos x gives:
−d + c − 6d = 0
i.e.
c − 7d = 0
(2)
Equating the coefficients of sin x gives:
−c − d − 6c = −50
i.e. − 7c − d = −50
(3)
Solving equations (2) and (3) gives: c = 7 and
d = 1.
Hence the P.I.,
υ = −2x −
1
3 + 7 sin x + cos x
(vi) The general solution, y = u +v,
i.e. y = Ae
2x
+ Be
−3x
− 2x
−
1
3 + 7 sin x + cos x
Problem 10. Solve the differential equation
d 2 y
dx 2 − 2
d y
dx
+ 2y = 3e x cos 2x, given that when
x = 0, y = 2 and
d y
dx
= 3.
Using the procedure of Section 51.2:
(i)
d 2 y
dx 2 − 2
d y
dx
+ 2y = 3e x cos 2x in D-operator
form is
(D
2
− 2D + 2)y = 3e
x cos 2x
(ii) The auxiliary equation is m 2 − 2m + 2 = 0
Using the quadratic formula,
m =
2 ±
√
[4 − 4(1)(2)]
2
=
2 ±
√
−4
2
=
2 ± j 2
2
i.e. m = 1 ± j 1.
(iii) Since the roots are complex, the C.F.,
u = e x (A cos x + B sin x).
(iv) Since the right hand side of the given differential equation is a product of an exponential and a cosine function, let the P.I.,
v = e x (C sin 2x + D cos 2x) (see Table 51.1(f) —
again, constants C and D are used since A and B
have already been used for the C.F.).
(v) Substituting v into (D 2 − 2D +2)v = 3e x cos 2x
gives:
(D
2
− 2D + 2)[e
x
(C sin 2x + D cos 2x)]
= 3e
x cos 2x
D(v) = e
x
(2C cos 2x − 2D sin 2x)
+ e
x
(C sin 2x + D cos 2x)
(≡e
x
{(2C + D) cos 2x
+ (C − 2D) sin 2x})
D
2
(v) = e
x
(−4C sin 2x − 4D cos 2x)
+ e
x
(2C cos 2x − 2D sin 2x)
+ e
x
(2C cos 2x − 2D sin 2x)
+ e
x
(C sin 2x + D cos 2x)
≡ e
x
{(−3C − 4D) sin 2x
+ (4C − 3D) cos 2x}
Hence (D 2 − 2D + 2)v
= e
x
{(−3C − 4D) sin 2x
+ (4C − 3D) cos 2x}
− 2e
x
{(2C + D) cos 2x
+ (C − 2D) sin 2x}
+ 2e
x
(C sin 2x + D cos 2x)
= 3e
x cos 2x