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Find two power series solutions of the given differential equation about the ordinary point x=0: y''-xy=0

Mathematics
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fourier series
.....
\[y=\sum_{n=0}^\infty c_nx^n\]

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Other answers:

\[y'=\sum_{n=1}^\infty c_nnx^{n-1}\]
\[y''=\sum_{n=2}^\infty c_n(n)(n-1)x^{n-2}\]
oh I need a review on this bookmark*
i'm pretty sure i have it down so you can come check it after \[y''-xy=\sum_{n=2}^\infty c_nn(n-1)x^{n-2}-x\sum_{n=0}^\infty c_nx^n\]
\[\sum_{n=2}^\infty c_nn(n-1)x^{n-2}-\sum_{n=0}^\infty c_nx^{n+1}\]
\[k=n-2\] \[k+2=n\] \[\sum_{k=0}^\infty c_{k+2}(k+2)(k+2-1)x^k-\sum_{n=0}^\infty c_nx^{n+1}=0\]
\[\sum_{k=0}^\infty c_{k+2}(k+2)(k+1)x^k-\sum_{k=1}^\infty c_{k-1}x^k\] where \[k=n+1\] \[k-1=n\]
pull out one term k=0 of first \[2c_2+\sum_{k=1}^\infty c_{k+2}(k+2)(k+1)x^k-\sum_{k=1}^\infty c_{k-1}x^k\]
\[2c_2+\sum_{k=1}^\infty [c_{k+2}(k+2)(k+1)-c_{k-1}]x^k\]
since identity 0=0 the sum and \[2c_2=0\] \[c_2=0\]
\[c_{k+2}(k+2)(k+1)=c_{k-1}\] \[c_{k+2}=\frac{c_{k-1}}{(k+2)(k+1)}\]
for k=1,2,3,4,5,6......
k=1 \[c_3=\frac{c_0}{3*2}\] k=2 \[c_4=\frac{c_1}{4*3}\] for k =3 \[c_5=\frac{c_2}{constant}\] =\[c_2=0\] k=4 \[c_6=\frac{c_3}{6*5)}=\frac{c_0}{6*5*3*2}\]
k=5 \[c_7=\frac{c_4}{7*6}=\frac{c_1}{7*6*4*3}\]
\[c_8=c_5=0\]
@TuringTest look correct so far?
like I said: *bookmark I want you to remind me; it's been a while... maybe @experimentX can verify better than me
jeez ... i forgot all those stuffs. i need to review it myself.
hm... @Zarkon care to verify a DE series solution?
Is zarkon the computer on lol
these standard equations power solutions look intimidating ...
\[y=c_0+c_1x+0+\frac{c_0}{3*2}x^3+\frac{c_1}{4*3}x^4+0+\frac{c_0}{2*3*5*6}x^6\] \[+\frac{c_1}{3*4*6*7}x^7+0......\]
\[y_1(x)=1+\frac{1}{2*3}x^3+\frac{1}{2*3*5*6}x^6...\] \[y_2(x)=x+\frac{1}{3*4}x^4+\frac{1}{3*4*6*7}x^7.......\]
what kind of function is this http://www.wolframalpha.com/input/?i=y%27%27+-+xy%3D0
Using Maple I found this solution \[ C0+C1*x+(1/6)*C0*x^3+(1/12)*C1*x^4+\\ (1/180)*C0*x^6+(1/504)*C1*x^7+O(x^8) \]
thats what i have
only i left it not multiplied so i can creat a summation
well, i guess you are right ... i never liked power series solution you know!!

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