Derive a seventh order power series representation of the general solution to Airy's equation $y^n(x)=xy(x)$. Initial conditions to the general solution $y(0)=1,$ $ y'(0)=1.$ (We don't need to write a general formula for the $ n-th$ term. Just write all the terms up to and including $x^7$).
I have done it but cannot write it as a power series of seventh order.
My attempt:I take as usual $y=\sum_{k=0}^{\infty} a_kx^k$ and then $y^n=\sum_{k=n}^{\infty} \frac{k!}{(k-n)!}a_kx^{k-n}$. And then rearranging the terms and using initial conditions we get $a_0=1$ and $a_1=1$ and $a_{k+n}=\frac{k!}{(k+n)!}a_{k-1}$ and obviously $a_n=0$. Then from here we get $a_{n+1}=\frac{1}{(n+1)!}a_0$$=$$\frac{1}{(n+1)!}$, $a_{n+2}=\frac{2}{(n+2)!}a_1$=$\frac{2}{(n+2)!}$, proceeding similarly. But according to the question we have to write upto and including $x^7$. Please help me to find the required power series representation as I failed it to form without using $a_n$.

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    $\begingroup$ Check for typos in your task, it should be $y''$, not $y^n$. This mistake happens from time to time one way or the other. See also other sources for Airy equation or Airy function. $\endgroup$ – LutzL Jul 13 '18 at 6:24
  • $\begingroup$ @LutzL. sir I am also surprised. But it $y^n$ there. $\endgroup$ – abcdmath Jul 13 '18 at 6:28
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    $\begingroup$ Then ask your tutor, it is probably a mistake in reading the hand-written original. You could write down your general solution and then add the numbers for $n=2$. $\endgroup$ – LutzL Jul 13 '18 at 7:12
  • $\begingroup$ The Airy equation is $y''=xy$ $\endgroup$ – Dylan Jul 13 '18 at 7:56
  • $\begingroup$ Maybe of interest math.stackexchange.com/questions/1197528/solution-of-yxy-0 and math.stackexchange.com/questions/567707/… $\endgroup$ – cgiovanardi Jul 13 '18 at 17:55

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