I'm reviewing Cauchy-Euler ODE's and am going to be trying to solve the following problem, don't remember much about these. Posted below momentarily.

Question

mendicant_bias · Answer

$$x^{2}y''-2y=0$$

mendicant_bias · Answer

If I remember correctly, the first thing we're supposed to do when solving this is solve the characteristic equation.

mendicant_bias · Answer

Is the characteristic equation supposed to be $$m^2 - 2m ?$$

mendicant_bias · Answer

@dan815

mendicant_bias · Answer

(Or, is this how I'm supposed to be approaching the problem?)

dan815 · Answer

u get the characteristic by assuming a solution is in form y=x^n

dan815 · Answer

$$y'=n*^{n-1}\
y''=n(n-1)*x^{n-2}\
subbing\
x^2y^{''}-2y=0\
x^2*n(n-1)*x^{n-2}-2*x^n=0\
n(n-1)*x^n-2*x^n=0\
x^n(n(n-1)-2)=0\
x^n=0,n(n-1)-2=0\
n^2-n-2=0\
(n+1)(n-2)=0; n=-1,2$$

dan815 · Answer

$$therfore~~ y=x^{-1}= 1/x , y=x^2$$

dan815 · Answer

Y_gen=c1/x+c2x^2

mendicant_bias · Answer

Alright, cool, totally forgot how to do this, lol. Looking at this, one second.

mendicant_bias · Answer

Alright, yeah, this makes sense; I don't entirely understand the math behind x^n equaling zero and then it disappearing, but I get how to do this (?).

Now I just need to remember how to do the rest of this. Now I find a particular solution?