504 Higher Engineering Mathematics
=
a r−1
2r 2 −
1
2
−
1
2
− r + 1
=
a r−1
2r 2 − r
=
a r−1
r(2r − 1)
Thus, when r = 1, a 1 =
a 0
1(2 − 1)
=
a 0
1 × 1
when r = 2, a 2 =
a 1
2(4 − 1)
=
a 1
(2 × 3)
=
a 0
(2 × 3)
when r = 3, a 3 =
a 2
3(6 − 1)
=
a 2
3 × 5
=
a 0
(2 × 3) × (3 × 5)
when r = 4, a 4 =
a 3
4(8 − 1)
=
a 3
4 × 7
=
a 0
(2×3×4)×(3×5×7)
and so on.
From equation (23), 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 =
1
2
and the above values of a 1 , a 2 ,
a 3 , … into the trial solution gives:
y=x
1
2
a 0 +a 0 x +
a 0
(2×3)
x
2
+
a 0
(2×3)×(3×5)
x
3
+
a 0
(2 × 3 × 4) × (3 × 5 × 7)
x
4
+ · · ·
i.e. y = a 0 x
1
2
1 + x +
x 2
(2 × 3)
+
x 3
(2 × 3) × (3 × 5)
+
x 4
(2 × 3 × 4) × (3 × 5 × 7)
+ · · ·
(27)
Since a 0 is an arbitrary (non-zero) constant in
each solution, its value could well be different.
Let a 0 = A in equation (26), and a 0 = B in equation (27). 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),
i.e. y = Ax
1 +
x
(1 × 3)
+
x 2
(1 × 2)× (3 × 5)
+
x 3
(1 × 2 × 3)× (3 × 5 ×7)
+
x 4
(1 × 2 × 3×4)×(3×5×7×9)
+· · ·
+ B x
1
2
1 + x +
x 2
(2 × 3)
+
x 3
(2 × 3)× (3 × 5)
+
x 4
(2 × 3 × 4) ×(3 × 5 ×7)
+· · ·
Problem 9. Use the Frobenius method to
determine the general power series solution of the
differential equation:
d 2 y
dx 2 − 2y = 0.
The differential equation may be rewritten as:
y − 2y = 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
+ · · ·
(28)
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
+ · · ·
(29)
(ii) Differentiating equation (29) 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) Replacing r by (r + 2) in
a r (c + r − 1)(c + r) x c+r−2 gives:
a r+2 (c + r + 1)(c + r + 2)x c+r
=
a r−1
2r 2 −
1
2
−
1
2
− r + 1
=
a r−1
2r 2 − r
=
a r−1
r(2r − 1)
Thus, when r = 1, a 1 =
a 0
1(2 − 1)
=
a 0
1 × 1
when r = 2, a 2 =
a 1
2(4 − 1)
=
a 1
(2 × 3)
=
a 0
(2 × 3)
when r = 3, a 3 =
a 2
3(6 − 1)
=
a 2
3 × 5
=
a 0
(2 × 3) × (3 × 5)
when r = 4, a 4 =
a 3
4(8 − 1)
=
a 3
4 × 7
=
a 0
(2×3×4)×(3×5×7)
and so on.
From equation (23), 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 =
1
2
and the above values of a 1 , a 2 ,
a 3 , … into the trial solution gives:
y=x
1
2
a 0 +a 0 x +
a 0
(2×3)
x
2
+
a 0
(2×3)×(3×5)
x
3
+
a 0
(2 × 3 × 4) × (3 × 5 × 7)
x
4
+ · · ·
i.e. y = a 0 x
1
2
1 + x +
x 2
(2 × 3)
+
x 3
(2 × 3) × (3 × 5)
+
x 4
(2 × 3 × 4) × (3 × 5 × 7)
+ · · ·
(27)
Since a 0 is an arbitrary (non-zero) constant in
each solution, its value could well be different.
Let a 0 = A in equation (26), and a 0 = B in equation (27). 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),
i.e. y = Ax
1 +
x
(1 × 3)
+
x 2
(1 × 2)× (3 × 5)
+
x 3
(1 × 2 × 3)× (3 × 5 ×7)
+
x 4
(1 × 2 × 3×4)×(3×5×7×9)
+· · ·
+ B x
1
2
1 + x +
x 2
(2 × 3)
+
x 3
(2 × 3)× (3 × 5)
+
x 4
(2 × 3 × 4) ×(3 × 5 ×7)
+· · ·
Problem 9. Use the Frobenius method to
determine the general power series solution of the
differential equation:
d 2 y
dx 2 − 2y = 0.
The differential equation may be rewritten as:
y − 2y = 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
+ · · ·
(28)
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
+ · · ·
(29)
(ii) Differentiating equation (29) 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) Replacing r by (r + 2) in
a r (c + r − 1)(c + r) x c+r−2 gives:
a r+2 (c + r + 1)(c + r + 2)x c+r