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"Given a differential equation of the form $a(x)y''+b(x)y'+y=0$, find functions $a(x)$ and $b(x)$ so that $y=x$ and $y=x^2$ are each a solution of this differential equation."

I'm really not sure how to approach this, at first I tried setting it up like a system of equations and tried to solve for $a(x)$ and $b(x)$ but it didn't work out. I'm really looking for a way to start this problem rather than an answer to it.

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1 Answer 1

up vote 6 down vote accepted

Since $y=x$ is a solution then it satisfies the DE so that $a(x) \cdot 0 + b(x)\cdot 1 +x=0 $ thus $b(x)=-x$. Similarly for $y=x^2$, we have $a(x) \cdot (2)+ (-x) \cdot(2x) +x^2=0$, which gives that $a(x)=\frac{1}{2}x^2$.

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