1. The problem statement, all variables and given/known data
Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of
(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0
Note that x=0 is an ordinary point.
2. Relevant equations
y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)
3. The attempt at a solution
(x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0
Making all the series start with the x3 term
2C2x0+[itex]\sum[/itex](k(k-1))ckxk (k=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](k(k-1))ckxk-2 (k=5 to ∞) + [itex]\sum[/itex](kckxk+2) (k=1 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]ckxk (k=3 to ∞)
2C2x0+[itex]\sum[/itex](n(n-1))cnxn (n=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](n+2)(n+1))cn+2xn (n=3 to ∞) + [itex]\sum[/itex]((n-2)cn-2xn) (n=3 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]cnxn (n=3 to ∞)
2C2x0-2C2x0-6C3x1-12C4x2+C0x0+C1x1+C2x2=0
c0=0
c3={c1-2}/{6}
c4={c2}/{12}
cn+2={n(n-1)cn+(n-2)cn-2+cn}/{(n+2)(n-1)} n=3,4,5...
I started with c5={(7c3+c1)}/{10}
but both of my c1 and c3 terms are tied up in the same equation, not really sure were to go from here
Let y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞) be a power series solution of
(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0
Note that x=0 is an ordinary point.
2. Relevant equations
y(x)=[itex]\sum[/itex]ckxk (k=0 to ∞)
y'(x)=[itex]\sum[/itex](kckxk-1) (k=1 to ∞)
y''(x)=[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞)
3. The attempt at a solution
(x2-1)[itex]\sum[/itex](k(k-1))ckxk-2 (k=2 to ∞) +[itex]\sum[/itex](kckxk) (k=1 to ∞)+[itex]\sum[/itex]ckxk (k=0 to ∞) -2x=0
Making all the series start with the x3 term
2C2x0+[itex]\sum[/itex](k(k-1))ckxk (k=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](k(k-1))ckxk-2 (k=5 to ∞) + [itex]\sum[/itex](kckxk+2) (k=1 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]ckxk (k=3 to ∞)
2C2x0+[itex]\sum[/itex](n(n-1))cnxn (n=3 to ∞)- 2C2x0-6C3x1-12C4x2-[itex]\sum[/itex](n+2)(n+1))cn+2xn (n=3 to ∞) + [itex]\sum[/itex]((n-2)cn-2xn) (n=3 to ∞)+C0x0+C1x1+C2x2+[itex]\sum[/itex]cnxn (n=3 to ∞)
2C2x0-2C2x0-6C3x1-12C4x2+C0x0+C1x1+C2x2=0
c0=0
c3={c1-2}/{6}
c4={c2}/{12}
cn+2={n(n-1)cn+(n-2)cn-2+cn}/{(n+2)(n-1)} n=3,4,5...
I started with c5={(7c3+c1)}/{10}
but both of my c1 and c3 terms are tied up in the same equation, not really sure were to go from here
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