# Find a polynomial f(x) that satisfies this equation: xf " (x) + f(x) = x^2

Find a polynomial f(x) that satisfies this equation:
$x{f}^{″}\left(x\right)+f\left(x\right)={x}^{2}$
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Look for a solution of the form:
$f\left(x\right)=\sum _{r=0}^{n}{a}_{r}{x}^{r}$
For some set of constants ${a}_{r}$
Then ${f}^{\prime }\left(x\right)=\sum _{r=1}^{n}{a}_{r}r{x}^{r-1}$
${f}^{″}\left(x\right)=\sum _{r=2}^{n}{a}_{r}r\left(r-1\right){x}^{r-2}$
So we have
$x\sum _{r=2}^{n}{a}_{r}r\left(r-1\right){x}^{r-2}+\sum _{r=0}^{n}{a}_{r}{x}^{r}={x}^{2}$
Equating like terms on lhs and rhs we see that:
$6{a}_{3}+{a}_{2}=1$ (for the ${x}^{2}$ terms)
$2{a}_{2}+{a}_{1}=0$ (for the x terms)
${a}_{0}=0$ (for the constant terms)
But we must have ${a}_{2}=1$ for the squared termsbecause there's just ${x}^{2}$ on rhs
Hence ${a}_{1}=-2$
${a}_{3}=0$
So $f\left(x\right)={x}^{2}-2x$