By "ck" I'm assuming you're referring to the series solution's coefficients, i.e. [tex]y=\displaystyle\sum_{k\ge0}c_kx^k[/tex].
Differentiating gives
[tex]y'=\displaystyle\sum_{k\ge1}kc_kx^{k-1}[/tex]
[tex]y''=\displaystyle\sum_{k\ge2}k(k-1)c_kx^{k-2}[/tex]
Substituting into the ODE yields
[tex]\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[/tex]
[tex]\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[/tex]
By extracting the first term ([tex]k=0[/tex]) of the third series, you can consolidate the second and third series:
[tex]\displaystyle\sum_{k\ge0}c_kx^{k+1}=\sum_{k\ge1}c_kx^{k+1}+c_0x[/tex]
Next, since the first series starts with a constant term while the other two start with a quadratic term, remove the first two terms:
[tex]\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[/tex]
Now shift the index so that the series begins at the same starting index by replacing [tex]k[/tex] with [tex]k+3[/tex].
[tex]\displaystyle\sum_{k\ge4}k(k-1)c_kx^{k-2}=\sum_{k\ge1}(k+3)(k+2)c_{k+3}x^{k+1}[/tex]
So now the series can be joined into one:
[tex]\displaystyle\sum_{k\ge1}\bigg[(k+3)(k+2)c_{k+3}-kc_k+c_k\bigg]x^{k+1}[/tex]
which means the ODE is equivalent to
[tex]\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[/tex]
It follows that [tex]2c_2=0\implies c_2=0[/tex] and [tex]c_0+6c_3=0[/tex]. Using this, you solve the recurrence
[tex](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[/tex]
with the initial values [tex]c_0[/tex] and [tex]c_1[/tex].
Because [tex]c_2=0[/tex], it follows that [tex]c_5=0,c_8=0,\ldots[/tex]; in general, [tex]c_{k=3\ell-1}=0[/tex] for all [tex]\ell\ge1[/tex].
Also, because the RHS of the recurrence vanishes when [tex]k=1[/tex], it follows that [tex]c_4=0,c_7=0,\ldots[/tex]; or generally, [tex]c_{k=3\ell+1}=0[/tex] for [tex]\ell\ge1[/tex], leaving you with only one not-necessarily-zero term of [tex]c_1[/tex].
So the hard part is finding a solution for [tex]c_k[/tex] when [tex]k[/tex] is a multiple of 3. By finding the first few of these terms, you'll start to notice a pattern. Starting with [tex]c_0[/tex], you find
[tex]c_3=-\dfrac1{3\times2}c_0[/tex]
[tex]c_6=\dfrac2{6\times5}c_3=-\dfrac1{6\times5\times3}c_0[/tex]
[tex]c_9=\dfrac5{9\times8}c_6=-\dfrac1{9\times8\times6\times3}c_0[/tex]
[tex]c_{12}=\dfrac9{12\times11}c_9=-\dfrac1{12\times11\times9\times6\times3}c_0[/tex]
and so on, following a general pattern of
[tex]c_{k=3\ell}=-\dfrac{c_0}{(3\ell-1)\prod\limits_{i=1}^\ell 3i}[/tex]
[tex]c_{k=3\ell}=-\dfrac{c_0}{3^\ell\ell!(3\ell-1)}[/tex]
valid for [tex]\ell\ge1[/tex].
So the series solution is
[tex]y=\displaystyle\sum_{k\ge0}c_kx^k[/tex]
[tex]y=c_0+c_1x+c_2x^2+c_3x^3+c_4x^4+c_5x^5+c_6x^6+\cdots[/tex]
[tex]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}[/tex]
[tex]y=c_0+c_1x+\displaystyle\sum_{\ell\ge1}c_{3\ell}x^{3\ell}[/tex]
[tex]y=c_0+c_1x-c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(3\ell-1)}[/tex]
[tex]y=c_0+c_1x+c_0\displaystyle\sum_{\ell\ge1}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}[/tex]
[tex]y=c_1x+c_0\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}[/tex]
So the two solutions are [tex]y_1=x[/tex] and [tex]y_2=\displaystyle\sum_{\ell\ge0}\frac{x^{3\ell}}{3^\ell\ell!(1-3\ell)}[/tex].
To solve the IVP, notice that when [tex]x=0[/tex], you have
[tex]y(0)=\displaystyle\sum_{k\ge0}c_kx^k=c_0=2[/tex]
[tex]y'(0)=\displaystyle\sum_{k\ge1}kc_kx^{k-1}=c_1=5[/tex]
so the solution to the IVP is the same as that to the general solution with the unknowns replaced accordingly.
