So I'm learning to solve ODE's with series on my own using Boyce and DiPrima and exercise #3 is irking me...just looking for power series solutions around the ordinary point... $$y''-xy'-y=0$$ So I get started with $$y=\sum_{k=0}^{\infty}a_kx^k, y'=\sum_{k=0}^{\infty}(k+1)a_{k+1}x^k, y''=\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^k$$ Now the first derivative term gets multiplied by $x$, so $$xy'=x\sum_{k=0}^{\infty}(k+1)a_{k+1}x^k=\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k+1}$$ To see the series I can expand it out and thus $$\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k+1}=a_1x+2a_2x^2+3a_3x^3+...$$ I need the summation in terms of $x^n$ so I can just rewrite as $$\sum_{k=0}^{\infty}(k+1)a_{k+1}x^{k+1}=\sum_{k=0}^{\infty}ka_kx^k$$ Now I can substitute all the derivative series into my ODE ad I get $$y''-xy'-y=\sum_{k=0}^{\infty}(k+2)(k+1)a_{k+2}x^k-\sum_{k=0}^{\infty}ka_kx^k-\sum_{k=0}^{\infty}a_kx^k$$ $$=\sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2}-ka_k-a_k]x^k=\sum_{k=0}^{\infty}[(k+2)(k+1)a_{k+2}-(k+1)a_k]x^k$$ So now to solve the ODE I need to first solve the recurrence, get the $a_k$ in terms of $a_0$ and $a_1$. I get $$a_{k+2}=\frac{a_k}{k+2}$$ To see a couple of terms, $$a_0=a_0, a_1=a_1, a_2=\frac{a_0}{2}, a_3=\frac{a_1}{3}, a_4=\frac{a_0}{2\cdot4}, a_5=\frac{a_1}{3\cdot5}, ...$$ So finally I'm ready to solve for y. Substituting in my $a_k$, $$y=a_0+\frac{a_1}{1}x+\frac{a_0}{2}x^2+\frac{a_1}{1 \cdot 3}x^3+\frac{a_0}{2 \cdot 4}x^4+\frac{a_1}{1 \cdot 3 \cdot 5}+...$$ $$=a_0\sum_{k=0}^{\infty}\frac{x^{2k}}{2^kk!}+a_1\sum_{k=0}^{\infty}\frac{2^kk!x^{2k+1}}{(2k+1)!}$$ I think this is the answer (the answer page got ripped out by my son...)
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$\begingroup$Everything's good exceptthe last expressions - those aren't quite factorials, they only use either the odd or even terms. They're what are called double factorials. For instance, 4! = 4x3x2x1, whereas 4!! = 4x2, and 5!! will be 5x3x1.
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