Please help for finding 2 Linearly Independent solutions of ....

Please help for finding 2 Linearly Independent solutions of

x^{2}y'' + xy' + 4y = 0

my method:

factoring first

let y=x^{A }so y' = Ax^{(A-1)} and y'' = A(A-1)x^{(A-2) }So A(A-1) + A + 4 =0

which is A^{2} + 4 = 0

A = +/- 2i

then I don't know how to find 2 LI solutions, please help.. (Doh)

Re: Please help for finding 2 Linearly Independent solutions of ....

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

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

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

By the principle of superposition, we then find:

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

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