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Find, using the power series: $$y(x)=\sum_{k=0}^\infty a_{k}x^k$$ a solution for the following differential equation: $$y'(x) = -x^2y(x),\,\, y(0)=1$$ What's the convergence radius of the constructed power series? Also give a closed formula.

So far I've come up with $\sum_{k=0}^\infty a_{k+1}(k+1)x^{k} = \sum_{k=0}^\infty -a_{k}x^{k+2}$. Usually I try to find a recursive formula for the coefficients and then try and guess a formula for the coefficients, but I dont know how to compare the coefficients on this one as the powers of x don't coincide. How should I go on from here, all I know is that $a_0=0$.

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  • $\begingroup$ Have you tried solving it by separating it? That should give you a good hint of what the power series is. $\log(y)=-\frac{1}{3}x^3 \Rightarrow y=e^{-1/3 x^3}$ I dont see how you could ever hope to come up with this using the power series approach you're doing, unless you know by heart the power series of the gaussian. What I just realized you could do is simply find the taylor expansion of $y$, using your initial values and the differential equation. $\endgroup$ – Ukhrir Jan 14 '14 at 20:45
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You can compare the coefficients here too, because $\sum_{k=0}^\infty -a_{k}x^{k+2} = \sum_{k=2}^\infty -a_{k-2}x^k$. (If you haven't seen this before, just write out the first few terms of each side to see why.)

So if we set $a_{-2} = a_{-1} = 0$, we get the recurrence $a_{k+1}(k+1) = -a_{k-2}$ for $k \ge 0$. The condition $y(0)=1$ gives $a_0=1$, and the remaining terms are now determined. It turns out that $a_k = 0$ unless $k$ is a multiple of $3$, so $y$ is a function of $x^3$. (More precisely: $y$ can be expressed as a power series in $x^3$.)

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Since $y$ has a power series representation, it is equal to its taylor series within its radius of convergence. Therefore, taking the first terms of the taylor series we see that $f(0)=1$, $f'(0)=0$, $f''(0)=(-x^2y(x))'=(-2xy(x)-x^2y'(x))(0)=0, $ and finally $f'''(0)=(-2xy(x)-x^2y'(x))(0)=-2$. We then merely have to show that all derivaties of order $1+3n$ are zero, as well as $2+3n$, which is easily done by induction based on the previous calculations. Further, we see that all derivatives $3+3n$ are of the appropriate form for $e^{-1/3 x^3}$ (which i must confess i didn't really feel like calculating).

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