Solution of x^2y''+xy'-y=0 and Regular Point of (1-x^2)y''-2xy'+2y=0

[latentcorpse]

What is the general solution of

$x^2y''+xy'-y=0$
i tried a series solution $y=\sum_{n=0}^{\infty} a_n x^n$
and whitteld it down to

$\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0$ but this isn't getting me anywhere






secondly how do i show that x=0 is a regular point of
$(1-x^2)y''-2xy'+2y=0$

the definition of a regular point is that :
x0 is a regular point of y''+a1(x)y'+a0(x)y=0 if the coefficients a1 and a0 can be expressed as convergent power series in (x-x0) for this value of x0.

so if i rearrange our equation into the desired form i get that

$a_1(x)=\frac{-2x}{1-x^2},a_2(x)=\frac{2}{1-x^2}$
for x0=0, a1=0 and a2=2.

what do i say to round of my argument. something like:

"clearly 0 and 2 can be expressed as a convergent power series in x" or
"clearly a1 and a0 arent singular for this value of x0"
how do i round this off?

 

[lanedance]

hi latentcorpse - does you series solution imply all $$a_n$$ are zero except for a certain n?

 

[HallsofIvy]

A series solution is not the best way to do this. Are you required to use series?

If so, look closely at lanedance's suggestion. If not, try y= x^m for some number m.

For your second question, remember that a geometric series, $\sum r^n$ has sum 1/(1- r).

 

[latentcorpse]

ok if i was just using series solution, i see now that $a_1$ will be non zero - is there any way of finding its value or is the solution just going to be $y=a_1x$?

i amen't required to use series solutions but it was a past paper exam question and i couldn't think of any other way to do it - why did you decide to use that substitution?



i don't see how the geometric sum fits into this last bit either.

 

[HallsofIvy]

You said you had $\sum_{n=0}^{\infty} (n^2-1) a_n x^n=0$.

For Every n, either $a_n= 0$ or $n^2- 1= 0$.

The reason I suggested x^m is that (x^m)'= mx^m-1 and (x^m)"= m(m-1)x^m-1. so that x(x^m)'= mx^m and x²(x^m)"= m(m-1)x^m so every term has the same power of x.

Finally, since the sum of the geometric series $\sum ar^n$ is $a/(1- r)$, $2/(1-x^2)$ can be written as $\sum 2(x^2)^n= \sum 2x^{2n}$. $2x/(1- x^2)$ is just that multiplied by x: $\sum 2x^{2n+1}$.