504 Higher Engineering Mathematics
=
ar− 1
2 r^2 −
1
2
−
1
2
−r+ 1
=
ar− 1
2 r^2 −r
=
ar− 1
r(2r−1)
Thus, whenr=1,a 1 =
a 0
1 ( 2 − 1 )
=
a 0
1 × 1
whenr=2,a 2 =
a 1
2 ( 4 − 1 )
=
a 1
( 2 × 3 )
=
a 0
( 2 × 3 )
whenr=3,a 3 =
a 2
3 ( 6 − 1 )
=
a 2
3 × 5
=
a 0
( 2 × 3 )×( 3 × 5 )
whenr=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=xc
{
a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···
+arxr+···
}
Substitutingc=
1
2
and the above values ofa 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)
Sincea 0 is an arbitrary (non-zero) constant in
each solution, its value could well be different.
Leta 0 =Ain equation (26), anda 0 =Bin equa-
tion (27). Also, if the first solution is denoted by
u(x)and the second byv(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)
+···
}
+Bx
(^12)
{
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
− 2 y=0.
The differential equation may be rewritten as:
y′′− 2 y=0.
(i) Let a trial solution be of the form
y=xc
{
a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···
+arxr+···
}
(28)
wherea 0 =0,
i.e.y=a 0 xc+a 1 xc+^1 +a 2 xc+^2 +a 3 xc+^3
- ···+arxc+r+··· (29)
(ii) Differentiating equation (29) gives:
y′=a 0 cxc−^1 +a 1 (c+ 1 )xc+a 2 (c+ 2 )xc+^1 - ···+ar(c+r)xc+r−^1 +···
andy′′=a 0 c(c− 1 )xc−^2 +a 1 c(c+ 1 )xc−^1
+a 2 (c+ 1 )(c+ 2 )xc+···
+ar(c+r− 1 )(c+r)xc+r−^2 +···
(iii) Replacingrby(r+ 2 )in
ar(c+r− 1 )(c+r)xc+r−^2 gives:
ar+ 2 (c+r+ 1 )(c+r+ 2 )xc+r