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}.