502 Higher Engineering Mathematics
Thus, when r = 1, a 2 =
a 1
(2 × 8)
=
a 0
(2 × 5 × 8)
since a 1 =
a 0
5
when r = 2, a 3 =
a 2
(3 × 11)
=
a 0
(2 × 3)(5 × 8 × 11)
when r = 3, a 4 =
a 3
(4 × 14)
=
a 0
(2×3×4)(5×8×11×14)
and so on.
From equation (16), the trial solution was:
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · ·+ a r x
r
+· · ·}
Substituting c =
2
3
and the above values of a 1 , a 2 ,
a 3 , … into the trial solution gives:
y = x
2
3
a 0 +
a 0
5
x +
a 0
2 × 5 × 8
x
2
+
a 0
(2 × 3)(5 × 8 × 11)
x
3
+
a 0
(2 × 3 × 4)(5 × 8 × 11 × 14)
x
4
+ · · ·
i.e. y = a 0 x
2
3
1 +
x
5
+
x 2
(2 × 5 × 8)
+
x 3
(2 × 3)(5 × 8 × 11)
+
x 4
(2 × 3 × 4)(5 × 8 × 11 × 14)
+ · · ·
(22)
Since a 0 is an arbitrary (non-zero) constant in each
solution, its value could well be different.
Let a 0 = A in equation (21), and a 0 = B in equation (22).
Also, if the first solution is denoted by u(x) and the
second by v(x), then the general solution of the given
differential equation is y = u(x) + v(x). Hence,
y = A
1 +x +
x
2
(2 × 4)
+
x
3
(2 × 3)(4 × 7)
+
x 4
(2 ×3 × 4)(4 × 7 × 10)
+· · ·
+ B x
2
3
1 +
x
5
+
x 2
(2 × 5 ×8)
+
x 3
(2 × 3)(5 × 8 × 11)
+
x 4
(2 × 3 × 4)(5 × 8 × 11 × 14)
+· · ·
Problem 8. Use the Frobenius method to
determine the general power series solution of the
differential equation:
2x 2 d 2 y
dx 2 − x
d y
dx
+ (1 − x)y = 0.
The differential equation may be rewritten as:
2x 2 y − x y + (1 − x)y = 0.
(i) Let a trial solution be of the form
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+ · · ·
+ a r x
r
+ · · ·}
(23)
where a 0 = 0,
i.e. y = a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·
(24)
(ii) Differentiating equation (24) gives:
y
= a 0 cx
c−1
+ a 1 (c + 1)x
c
+ a 2 (c + 2)x
c+1
+ · · ·+ a r (c + r)x
c+r−1
+ · · ·
and y
= a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
(iii) Substituting y, y and y into each term of
the given equation 2x 2 y − x y + (1 − x)y = 0
gives:
2x
2 y
= 2a 0 c(c − 1)x
c
+ 2a 1 c(c + 1)x
c+1
+ 2a 2 (c + 1)(c + 2)x
c+2
+ · · ·
+ 2a r (c + r − 1)(c + r)x
c+r
+ · · ·
(a)
−x y
= −a 0 cx
c
− a 1 (c + 1)x
c+1
− a 2 (c + 2)x
c+2
− · · ·
− a r (c + r)x
c+r
− · · ·
(b)
(1 − x)y = (1 − x)(a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·)
= a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·
Thus, when r = 1, a 2 =
a 1
(2 × 8)
=
a 0
(2 × 5 × 8)
since a 1 =
a 0
5
when r = 2, a 3 =
a 2
(3 × 11)
=
a 0
(2 × 3)(5 × 8 × 11)
when r = 3, a 4 =
a 3
(4 × 14)
=
a 0
(2×3×4)(5×8×11×14)
and so on.
From equation (16), the trial solution was:
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+· · ·+ a r x
r
+· · ·}
Substituting c =
2
3
and the above values of a 1 , a 2 ,
a 3 , … into the trial solution gives:
y = x
2
3
a 0 +
a 0
5
x +
a 0
2 × 5 × 8
x
2
+
a 0
(2 × 3)(5 × 8 × 11)
x
3
+
a 0
(2 × 3 × 4)(5 × 8 × 11 × 14)
x
4
+ · · ·
i.e. y = a 0 x
2
3
1 +
x
5
+
x 2
(2 × 5 × 8)
+
x 3
(2 × 3)(5 × 8 × 11)
+
x 4
(2 × 3 × 4)(5 × 8 × 11 × 14)
+ · · ·
(22)
Since a 0 is an arbitrary (non-zero) constant in each
solution, its value could well be different.
Let a 0 = A in equation (21), and a 0 = B in equation (22).
Also, if the first solution is denoted by u(x) and the
second by v(x), then the general solution of the given
differential equation is y = u(x) + v(x). Hence,
y = A
1 +x +
x
2
(2 × 4)
+
x
3
(2 × 3)(4 × 7)
+
x 4
(2 ×3 × 4)(4 × 7 × 10)
+· · ·
+ B x
2
3
1 +
x
5
+
x 2
(2 × 5 ×8)
+
x 3
(2 × 3)(5 × 8 × 11)
+
x 4
(2 × 3 × 4)(5 × 8 × 11 × 14)
+· · ·
Problem 8. Use the Frobenius method to
determine the general power series solution of the
differential equation:
2x 2 d 2 y
dx 2 − x
d y
dx
+ (1 − x)y = 0.
The differential equation may be rewritten as:
2x 2 y − x y + (1 − x)y = 0.
(i) Let a trial solution be of the form
y = x
c
{a 0 + a 1 x + a 2 x
2
+ a 3 x
3
+ · · ·
+ a r x
r
+ · · ·}
(23)
where a 0 = 0,
i.e. y = a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·
(24)
(ii) Differentiating equation (24) gives:
y
= a 0 cx
c−1
+ a 1 (c + 1)x
c
+ a 2 (c + 2)x
c+1
+ · · ·+ a r (c + r)x
c+r−1
+ · · ·
and y
= a 0 c(c − 1)x
c−2
+ a 1 c(c + 1)x
c−1
+ a 2 (c + 1)(c + 2)x
c
+ · · ·
+ a r (c + r − 1)(c + r)x
c+r−2
+ · · ·
(iii) Substituting y, y and y into each term of
the given equation 2x 2 y − x y + (1 − x)y = 0
gives:
2x
2 y
= 2a 0 c(c − 1)x
c
+ 2a 1 c(c + 1)x
c+1
+ 2a 2 (c + 1)(c + 2)x
c+2
+ · · ·
+ 2a r (c + r − 1)(c + r)x
c+r
+ · · ·
(a)
−x y
= −a 0 cx
c
− a 1 (c + 1)x
c+1
− a 2 (c + 2)x
c+2
− · · ·
− a r (c + r)x
c+r
− · · ·
(b)
(1 − x)y = (1 − x)(a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·)
= a 0 x
c
+ a 1 x
c+1
+ a 2 x
c+2
+ a 3 x
c+3
+ · · · + a r x
c+r
+ · · ·