POWER SERIES METHODS OF SOLVING ORDINARY DIFFERENTIAL EQUATIONS 499I
the value ofcis obtained. From equation (18),sincea 0 =0, thenc= 0 orc=2
3(a) Whenc=0:
From equation (19), ifc=0,a 1 (1×1)−a 0 =0,
i.e.a 1 =a 0
From equation (20), ifc=0,
ar+ 1 (r+1)(3r+1)−ar=0,
i.e.ar+ 1 =
ar
(r+ 1 )( 3 r+ 1 )r≥ 0Thus, whenr=1,a 2 =
a 1
(2×4)=a 0
(2×4)
sincea 1 =a 0whenr=2,a 3 =a 2
(3×7)=a 0
(2×4)(3×7)ora 0
(2×3)(4×7)whenr=3,a 4 =a 3
(4×10)=a 0
(2× 3 ×4)(4× 7 ×10)
and so on.From equation (16), the trial solution was:
y=xc{a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···+arxr+···}
Substitutingc=0 and the above values ofa 1 ,a 2 ,
a 3 , ... into the trial solution gives:
y=x^0{
a 0 +a 0 x+(
a 0
(2×4))
x^2+(
a 0
(2×3)(4×7))
x^3+(
a 0
(2× 3 ×4)(4× 7 ×10))
x^4 +···}i.e. y=a 0
{
1 +x+x^2
( 2 × 4 )+x^3
( 2 × 3 )( 4 × 7 )+x^4
( 2 × 3 × 4 )( 4 × 7 × 10 )+···}
(21)(b) Whenc=
2
3:From equation (19), ifc=
2
3,a 1 (3)(
5
3)
−a 0 =0,i.e.a 1 =
a 0
5From equation (20), ifc=2
3
ar+ 1(
2
3+r+ 1)
(2+ 3 r+1)−ar=0,i.e.ar+ 1(
r+5
3)
(3r+3)−ar=ar+ 1 (3r^2 + 8 r+5)−ar=0,i.e.ar+ 1 =ar
(r+ 1 )( 3 r+ 5 )r≥ 0Thus, whenr=1, a 2 =a 1
(2×8)=a 0
(2× 5 ×8)sincea 1 =a 0
5whenr=2, a 3 =a 2
(3×11)=a 0
(2×3)(5× 8 ×11)
whenr=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=xc{a 0 +a 1 x+a 2 x^2 +a 3 x^3 +···+arxr+···}Substitutingc=2
3and the above values ofa 1 ,a 2 ,
a 3 , ... into the trial solution gives:y=x2
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 x2
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)Sincea 0 is an arbitrary (non-zero) constant in each
solution, its value could well be different.