Answer
See below
Work Step by Step
Given: $4x^2y''+xe^xy'-y=0$
We can notice that $p=\frac{e^x}{4}, q=-\frac{1}{4}$
and $p(0)=\frac{1}{4}, q(0)=-\frac{1}{4}$
Obtain: $r(r-1)+\frac{1}{4}r-\frac{1}{4}=0\\
4(r^2-r)+r-1=0\\
4r^2-3r-1=0\\
r=1, r=-\frac{1}{4}$
