
find the values of m
ok i wanted to know if someone could check my work and let me know if they got the same thing.
im supposed to find the values of m so that the function y = x^m is a solution to,
x^2y''  3xy'  y = 0.
i got m = 0. i just took the first and 2nd derivative of the function and plugged it into the equation and solved.
thanks in advance.

You should get : $\displaystyle m=2 \pm \sqrt{5}$.
so that the two solutions are : $\displaystyle y_1=x^{2+\sqrt{5}}$ & $\displaystyle y_2=x^{2\sqrt{5}}$.
Hence, the general solution is $\displaystyle y=c_1x^{2+\sqrt{5}}+c_2x^{2\sqrt{5}}$.

yeah i was or am a little tired, ive been trying to get ready for this exam all night..lol thanks for your reply.

Generalization :
In case we don't know there are solutions of the form $\displaystyle y=x^m$ and considering that we have an Euler equation, the subtitution $\displaystyle x=e^u$ with $\displaystyle u$ independent variable, transforms the given equation into a linear homogeneous differential equation with constant coefficients.
Fernando Revilla