# Please verify my power series solution to the differential equation $y''-xy'+y=0$.

Please verify my answer to the following differential equation: $$y''-xy'+y=0$$ Let $y = {\sum_{n=0}^\infty}C_nx^n$, then $y' = {\sum_{n=1}^\infty}nC_nx^{n-1}$ and $y''={\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}$

Substituting this to the equation we get

$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-x{\sum_{n=1}^\infty}nC_nx^{n-1}+{\sum_{n=0}^\infty}C_nx^n = 0$$ $${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$

Getting the $x^n$ term on all the terms $${\sum_{n=0}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$

Getting the $0$th term from the first and the third summations we get $$2C_2+C_0 + {\sum_{n=1}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=1}^\infty}C_nx^n = 0$$

Factoring $x^n$ we get $$2C_2+C_0 + {\sum_{n=0}^\infty}[(n+2)(n+1)C_{n+2}-nC_n+C_n]x^n= 0$$

i.$$2C_2+C_0 = 0 => C_2 = \frac{-C_0}{2}$$

ii.$$(n+2)(n+1)C_{n+2}-nC_n+C_n = 0$$

Therefore solving ii. for $C_{n+2}$ $$C_{n+2}=\frac{(n-1)C_n}{(n+2)(n+1)}, n=0,1,2,3,...$$

If $n = 0$,

$$C_2 = \frac{-C_0}{2!}$$

If $n=1$,

$$C_3 = 0$$

If $n=2$,

$$C_4 = \frac{C_2}{3*4} = \frac{-C_0}{4!}$$

If $n=3$,

$$C_5 = \frac{2C_3}{4*5}=0$$

If $n=4$, $$C_6 = \frac{3C_4}{5*6} = \frac{-C_0}{6!}$$

Upon seeing the pattern we realize that if $n=2m$ then $$C_{2m} = \frac{-C_0}{2m!}$$

And if $n=2m+1$ then $$C_{2m+1} = 0$$

So the final answer would be $$y = {\sum_{n=0}^\infty}C_nx^n => {\sum_{m=0}^\infty}\frac{-C_0*x^{2m}}{2m!}$$

• $$C_{n+2}=\frac{(n-1)C_n}{(n+2)(n+1)}, n=0,1,2,3,...$$ – socrates Apr 27 '17 at 1:25
• @MyGlasses When n = 1, $C_3 = 0$ – socrates Apr 27 '17 at 1:26
• right. you must have two solutions! – Nosrati Apr 27 '17 at 1:29
• The other solution is $0$. Is it not? @MyGlasses – socrates Apr 27 '17 at 1:30
• @MyGlasses Hello? – socrates Apr 27 '17 at 1:46

Your equation is a second order linear equation so it should be a two-dimensional space of solutions (that is, solutions that depend on two free parameters) while your final answer depends only on one free parameter $C_0$ so this means you did something wrong.

Indeed, your equations don't determine what is $C_1$ which actually means that $C_1$ can be arbitrary. In addition, you made a mistake in deducing the general pattern. For example,

$$C_6 = \frac{3}{6 \cdot 5} C_4 = -\frac{3}{6 \cdot 5} \frac{1}{4!} C_0 = -\frac{3C_0}{6!} \neq -\frac{C_0}{6!}.$$

In fact, we have

$$C_{2m} = \frac{(2m-3)C_{2m-2}}{(2m)(2m-1)} = \frac{(2m-3)(2m-5)C_{2m-4}}{(2m)(2m-1)(2m-2)(2m-3)} = \dots = -\frac{(2m - 3)(2m - 5) \dots 1}{(2m)!} C_0 = - \frac{(2m - 3)(2m - 5) \dots 1}{(2m)(2m-1)(2m-2)(2m-3)(2m-4) \dots 1} C_0 = -\frac{C_0}{(2m-1)2^{m}m!}$$

and the general solution is given by

$$y(x) = C_1 \cdot x - C_0 \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m-1)2^m m!}.$$

• How is the first solution $x$? – socrates Apr 27 '17 at 1:58
• @socrates: Because $C_1$ is free (while $C_3 = C_5 = \dots = 0$). You can also plug in $y = x$ into the equation and see that it solves it. You get two linearly independent solutions, $y(x) = x$ which is odd and $y(x) = \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m-1)2^m m!}$ which is even. – levap Apr 27 '17 at 2:00
• MyGlasses: I'm not sure what you mean but I had the summation running over $n$ while the summand depended only on $m$. I've fixed that so now it should make sense. Thanks! – levap Apr 27 '17 at 2:02
• If m = 3 then $C_6 = \frac{-C_0}{5*2^3*3!}$ – socrates Apr 27 '17 at 2:08
• @socrates: Yes, which is the same as $-\frac{3C_0}{6!}$. – levap Apr 27 '17 at 2:08

You should be able to do a change of variable to transform this equation into Hermite's equation then compare the resulting coefficients to Hermite's.

Alternatively, you can determine the Rodrigue's Operator formulation of solutions of the equation and verify the operators have the desired effect on your series.