Question

Given y'' - x^2y' + xy = 0 about (0,0)

Answer 1

By "ck" I'm assuming you're referring to the series solution's coefficients, i.e. $y=\displaystyle\sum_{k\ge0}c_kx^k$.

Differentiating gives

$y'=\displaystyle\sum_{k\ge1}kc_kx^{k-1}$
$y''=\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}$

Substituting into the ODE yields

$\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}-x^2\sum_{k\ge1}kc_kx^{k-1}+x\sum_{k\ge0}c_kx^k=0$
$\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}-\sum_{k\ge1}kc_kx^{k+1}+\sum_{k\ge0}c_kx^{k+1}=0$

By extracting the first term ($k=0$) of the third series, you can consolidate the second and third series:

$\displaystyle\sum_{k\ge0}c_kx^{k+1}=\sum_{k\ge1}c_kx^{k+1}+c_0x$

Next, since the first series starts with a constant term while the other two start with a quadratic term, remove the first two terms:

$\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}=\sum_{k\ge4}k(k-1)c_kx^{k-2}+2c_2+6c_3x$

Now shift the index so that the series begins at the same starting index by replacing $k$ with $k+3$.

$\displaystyle\sum_{k\ge4}k(k-1)c_kx^{k-2}=\sum_{k\ge1}(k+3)(k+2)c_{k+3}x^{k+1}$

So now the series can be joined into one:

$\displaystyle\sum_{k\ge1}\bigg[(k+3)(k+2)c_{k+3}-kc_k+c_k\bigg]x^{k+1}$

which means the ODE is equivalent to

$\displaystyle\sum_{k\ge1}\bigg[(k+3)(k+2)c_{k+3}-(k-1)c_k\bigg]x^{k+1}+c_0x+2c_2+6c_3x=0$

It follows that $2c_2=0\implies c_2=0$ and $c_0+6c_3=0$. Using this, you solve the recurrence

$(k+3)(k+2)c_{k+3}-(k-1)c_k=0\iff c_{k+3}=\dfrac{k-1}{(k+3)(k+2)}c_k$

with the initial values $c_0$ and $c_1$.

Because $c_2=0$, it follows that $c_5=0,c_8=0,\ldots$; in general, $c_{k=3\ell-1}=0$ for all $\ell\ge1$.

Also, because the RHS of the recurrence vanishes when $k=1$, it follows that $c_4=0,c_7=0,\ldots$; or generally, $c_{k=3\ell+1}=0$ for $\ell\ge1$, leaving you with only one not-necessarily-zero term of $c_1$.

So the hard part is finding a solution for $c_k$ when $k$ is a multiple of 3. By finding the first few of these terms, you'll start to notice a pattern. Starting with $c_0$, you find

$c_3=-\dfrac1{3\times2}c_0$
$c_6=\dfrac2{6\times5}c_3=-\dfrac1{6\times5\times3}c_0$
$c_9=\dfrac5{9\times8}c_6=-\dfrac1{9\times8\times6\times3}c_0$
$c_{12}=\dfrac9{12\times11}c_9=-\dfrac1{12\times11\times9\times6\times3}c_0$

and so on, following a general pattern of

$c_{k=3\ell}=-\dfrac{c_0}{(3\ell-1)\prod\limits_{i=1}^\ell 3i}$
$c_{k=3\ell}=-\dfrac{c_0}{3^\ell\ell!(3\ell-1)}$

valid for $\ell\ge1$.

So the series solution is

$y=\displaystyle\sum_{k\ge0}c_kx^k$
$y=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6+\cdots$
$y=c_0+c_1x+\displaystyle\sum_{\ell\ge1}c_{3\ell-1}x^{3\ell-1}+\sum_{\ell\ge1}c_{3\ell}x^{3\ell}+\sum_{\ell\ge1}c_{3k+1}x^{3k+1}$
$y=c_0+c_1x+\displaystyle\sum_{\ell\ge1}c_{3\ell}x^{3\ell}$
$y=c_0+c_1x-c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(3\ell-1)}$
$y=c_0+c_1x+c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}$
$y=c_1x+c_0\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}$

So the two solutions are $y_1=x$ and $y_2=\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}$.

To solve the IVP, notice that when $x=0$, you have

$y(0)=\displaystyle\sum_{k\ge0}c_kx^k=c_0=2$
$y'(0)=\displaystyle\sum_{k\ge1}kc_kx^{k-1}=c_1=5$

so the solution to the IVP is the same as that to the general solution with the unknowns replaced accordingly.