Can anybone help me solving this 2nd order DE?

Question

anonymous · Answer

$$x^2 y'' - 2xy' + 2y =0$$ if we know there has a solution in the form of $$y=x^a, a \in \mathbb{R} $$

loser66 · Answer

to me, $$y'=ax^{a-1}\y''=a(a-1)x^{a-2}$$
then replace to the original one to get $$x^2a(a-1)x^{a-2}-2xax^{a-1}+2x^a=0\a(a-1)x^a-2ax^a+2x^a=0$$
factor x^a out to get $$x^a(a^2-4a+1)=0$$
then, solve for a , reject x =0. then, plug a back to y =x^a

loser66 · Answer

However, you should wait for other, I didn't take differential equation yet. I just saw someone in this site solved this problem and learned from him. I am not sure about the way I solve or the result I get. :)

anonymous · Answer

Thank you very much for your help!

loser66 · Answer

hey, I told you have to wait for someone who masters the material!! don't trust me.

anonymous · Answer

Ok! I just wanted to thank you for your response!

loser66 · Answer

@FutureMathProfessor hey, you can check mine to help him/her, please

loser66 · Answer

@Callisto

anonymous · Answer

Give me a sec

amistre64 · Answer

your y,y',y'' is fine and your setup was good

loser66 · Answer

sorry for factor part, it should be $$x^a(a^2-4a+3)=0$$

anonymous · Answer

$$x^2y''-2xy'+2y=0$$If we know the solution is of the form $y=x^a$, recall that it only follows $y'=ax^{a-1},y''=a(a-1)x^{a-2}$ by the product rule. Substitute these into our expression and try to solve for $a$:$$a(a-1)x^a-2ax^a+2x^a=0$$Factor out $x^a$ to yield:$$x^a(a(a-1)-2a+2)=0$$It's clear the trivial solution $x=0\implies y=0$ works from the above, but that's obvious. Let's assume $x\ne0\implies x^a\ne0$ and we find:$$a(a-1)-2a+2=0\a(a-1)-2(a-1)=0$a-1)(a-2)=0$$... so it follows \(a=1$ and $a=2$ both satisfy our equation, yielding the solutions $y=x$ and $y=x^2$.

anonymous · Answer

Note this is a homogeneous linear second-order differential equation; the fact it is linear allows us to use the principle of superposition to deduce that *all* solutions to our equation exist are a vector subspace spanned by our two fundamental solutions found above -- i.e. any linear combination of the two solutions yields *another* solution, letting us write our general solution in the form$$y=Ax+Bx^2$$for all real $A,B$.

anonymous · Answer

It can be solved by the reduction of order method