Determine the values of $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$ given that $\displaystyle y=ax^2+bx+c$ is a solution to $\displaystyle x^2(y'')^2+y^2=80x^4+72x^3+50x^2+24x+16$ given the values $\displaystyle a$,$\displaystyle b$ and $\displaystyle c $ are positive.

$\displaystyle y'=2ax+b$ and $\displaystyle y''=2a$ hence $\displaystyle a^2x^4+2abx^3+(2ac+b^s+4a^2)x^2+c^2=80x^4+72x^3+50 x^2+24x+16$

However, from these results, I am not sure exactly how to determine which values you would use. I figured $\displaystyle a=\sqrt{80}$ and $\displaystyle c=4$, however, you get more than one option for $\displaystyle b$. Any help is greatly appreciated!