Thread:

1. 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.

2. You should get : $m=2 \pm \sqrt{5}$.

so that the two solutions are : $y_1=x^{2+\sqrt{5}}$ & $y_2=x^{2-\sqrt{5}}$.

Hence, the general solution is $y=c_1x^{2+\sqrt{5}}+c_2x^{2-\sqrt{5}}$.

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

4. Generalization :

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

Fernando Revilla
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