- #1
fluidistic
Gold Member
- 3,924
- 261
Homework Statement
Solve [itex]x^2y''-3y=x^3[/itex].
Show that there are many solutions [itex]\phi[/itex] such that [itex]\phi (0)= \phi '(0)=0[/itex].
Homework Equations
Not sure.
The Attempt at a Solution
It's a Cauchy-Euler equation so that I made the ansatz [itex]\phi (x)=x^\alpha[/itex]. I reached that [itex]x^\alpha [\alpha (\alpha -1 )-3]=x^3[/itex]. For this to be true for all x, I think that [itex]\alpha (\alpha -1 )-3[/itex] must equal [itex]0[/itex].
If this is right, I reach that [itex]\alpha = \frac{1 \pm \sqrt { 13}}{2}[/itex]. Because these 2 roots are real and distinct, I have that [itex]y(x)=c_1 x ^{ \frac{1+\sqrt 13}{2} }+c_2 x ^{ \frac{1-\sqrt 13}{2} }[/itex]. However this does not agree with Wolfram alpha: http://www.wolframalpha.com/input/?i=x^2y''-3y=x^3.
I don't know what I'm missing/did wrong. Any help is appreciated!