# Show that $e^{-x^2}$ is the solution to the initial value problme $y''+(2-4x^2)y=0$ $y(o)=1$ $y'(0)=0$

I have already worked out a lot of stuff already from previous parts of the questions and I think I've got to work from the fact that $a_{2k+1}=0$ and $a_{2k}= (-1)^k/k!$ for k>0 satisfies the recurrence relation of the power series as it states this and then says hence deduce that $y(x)=e^{-x^2}$

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## 2 Answers

$$e^x = \sum^{+\infty}_{k=0}\cfrac {x^k}{k!} \implies e^{-x^2}=\sum^{+\infty}_{k=0}\cfrac {(-x^2)^k}{k!}=\sum^{+\infty}_{k=0}\cfrac {(-1)^k}{k!}x^{2k}$$ The coefficient of $x^{2k}$ satisfies the sequence you found.

Working forward

$$\sum^{+\infty}_{k=0}a_k\ x^k =\sum^{+\infty}_{k=0}a_{2k}\ x^{2k}+\sum^{+\infty}_{k=0}a_{2k+1}\ x^{2k+1}=\sum^{+\infty}_{k=0}\cfrac {(-1)^k}{k!}\ x^{2k}=\sum^{+\infty}_{k=0}\cfrac {(-x^2)^k}{k!}=e^{-x^2}$$

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The only thing you have to do is to show that $f(x)=e^{-x^2}$ satisfies the equation with the initial values. By the existence and uniqueness theorem for ode, you will know that this is the solution.

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