x2y'' + xy' + 4y = 0

my method:

factoring first
let y=xA so y' = Ax(A-1) and y'' = A(A-1)x(A-2)

So A(A-1) + A + 4 =0
which is A2 + 4 = 0
A = +/- 2i

One way to proceed would be to use Euler's formula:

$y_1=c_1x^{2i}=c_1e^{\ln(x^{2i})}=c_1e^{2\ln(x)i}=c _1(\cos(2\ln(x))+i\sin(2\ln(x)))$

$y_2=c_2x^{-2i}=c_2e^{\ln(x^{-2i})}=c_2e^{-2\ln(x)i}=c_2(\cos(2\ln(x))-i\sin(2\ln(x)))$

By the principle of superposition, we then find:

$y_1=c_1\cos(2\ln(x))$

$y_2=c_2\sin(2\ln(x))$