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Could someone help me solve the initial value problem

$y'' - xy = x^2 \\ y(0) = 2 ; y'(0) = 1$

by the $\bf method\bf$ $\bf of\bf$ $\bf power\bf$ $\bf series \bf $.

I'm not really sure how to handle the $x^2$ when it isn't being multiplied by some form of $y$. Advice?

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    I suggest considering all the terms corresponding to powers of $x$ less than 3 individually and grouping the rest. What do you get when you try to substitute $y=\sum a_n x^n$?2012-03-18
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    This doesn't match the initial conditions, but I noticed that y = -x is a solution to the equation. Not sure if this means much.2012-03-18

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