Question

Find the general solution of x^2*y"+4xy'+2y=0 that is valid in any interval not including the singular point.

Answer 1

sorry , I have to mark it to be back (in case)

Answer 2

It's okay.

Answer 3

$$x^2y''+4xy'+2y=0\\y''+\frac4xy'+\frac2{x^2}y=0$$As a general guess, let's assume a solution of the form \(y=1/x^n\) and observe \(y'=-n/x^{n+1},y''=n(n+1)/x^{n+2}\):$$\frac{n(n+1)}{x^{n+2}}-\frac{4n}{x^{n+2}}+\frac2{x^{n+2}}=0\\\frac1{x^{n+2}}\left(n(n+1)-4n+2\right)=0\\n^2+n-4n+2=0\\n^2-3n+2=0\\(n-2)(n-1)=0\\n=1,2$$

Answer 4

Hence we observe \(\left\{\dfrac1x,\dfrac1{x^2}\right\}\) are our fundamental solutions (away from the singular point)

Answer 5

But how did you get y=1/x^n?

Answer 6

and thus our general solution is of the form \(y=\dfrac{c_1}x+\dfrac{c_2}{x^2}\)

Answer 7

@Idealist it's an educated guess to be honest; it's just like how you make a general guess of the form \(y=e^{\lambda t}\) for solving constant-coefficient linear equations. Here you should observe that our derivatives appear to lose powers of \(x\) (hence the \(x^2y'',xy',y\) for the terms to all cancel out and yield \(0\)) so it's only natural. The fact that we have a singular point suggested it might have poles so I tried the form \(1/x^n\). We know we have all of our solutions because this is a second-order linear equation and so our solution space is two-dimensional (hence two independent fundamental solutions)

Answer 8

Thanks, oldrin.

Answer 9

no problem :-) hope you understood

Answer 10

by the way, this is a Cauchy-Euler equation now that I think about it. A similar approach to the one I used above is described in the trial solution section: http://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation

Answer 11

to me, I solve it as usual
the characteristic equation is x^2 r^2 +4xr +2 =0 give me 2 roots
\[r = \frac {-4x\pm \sqrt{16x^2 -4(x^2)(2)}}{2x^2}\]
\[r = \frac{-2\pm\sqrt{2}}{x}\]
so the general solution of the equation is
\[\huge y = C_1 e ^{(\frac{-2+\sqrt{2}}{x})x} + C_2 e^ {(\frac{-2-\sqrt{2}}{2})x}\]
after cancel out the x , we don't have variable in the solution, so, it is valid at any interval.

Answer 12

but not sure whether it's right or wrong, just try.

Answer 13

@Loser66 unfortunately that solution does not work :/ the flaw is in assuming you can use a guess of that form despite this not being a constant-coefficient problem

Answer 14

got you, thanks for pointing it out. I didn't know it.

Answer 15

@Loser66 if there was a solution of the form \(e^{rt}\), however, then you would be correct.

Answer 16

oldrin, the website was very helpful. Thanks a lot. Have a good day.