anonymous
  • anonymous
Can anybone help me solving this 2nd order DE?
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[x^2 y'' - 2xy' + 2y =0\] if we know there has a solution in the form of \[y=x^a, a \in \mathbb{R} \]
Loser66
  • Loser66
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
  • Loser66
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. :)

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anonymous
  • anonymous
Thank you very much for your help!
Loser66
  • Loser66
hey, I told you have to wait for someone who masters the material!! don't trust me.
anonymous
  • anonymous
Ok! I just wanted to thank you for your response!
Loser66
  • Loser66
@FutureMathProfessor hey, you can check mine to help him/her, please
Loser66
  • Loser66
@Callisto
anonymous
  • anonymous
Give me a sec
amistre64
  • amistre64
your y,y',y'' is fine and your setup was good
Loser66
  • Loser66
sorry for factor part, it should be \[x^a(a^2-4a+3)=0\]
anonymous
  • anonymous
$$x^2y''-2xy'+2y=0$$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
  • anonymous
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
  • anonymous
It can be solved by the reduction of order method

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