Question

Find a polynomial f(x) satisfying the equation xf''(x) + 3f(x) = 4x^2.

Answer 1

We're going to assume that the solution is a quadratic polynomial, since we have the sum of the function and its derivative adds up to $4x^2$. So let $f(x)=ax^2+bx+c \implies f'(x)=2ax+b \text{ and } f''(x)=2a.$
Plug all these into the given DE to get

$2ax+3ax^2+3bx+3c=4x^2$ or
$3ax^2+(2a+3b)x+3c=4x^2$

From the above equation the left hand side should have a coffienet of 4 for $x^2$, and all other coffienents must be 0. That is
$3a=4,2a+3b=0$ and $3c=0$.

You can easily solve the system above and find $a=\frac{4}{3},b=-\frac{8}{9}$ and $c=0$.

Thus $f(x)=\frac{4}{3}x^2-\frac{8}{9}x$