Question

I’m stuck finding 2 linearly independent Frobenius series solutions of the differential equation:
2xy′′−y′−y=0
. I’ve worked out y_2 but I end up dividing by 0 when I try to work out y_1 via my reccurence relation:
a_n=a_(n−1)/((n−1)(2n−3))
. Can someone please give me a hint with this.

Answer 1

y = sum(0) a_n x^(n)
y'= sum(1) a_n n x^(n-1)
y''= sum(2) a_n n(n-1) x^(n-2)

sub in and line up exponents and indexes

2x sum(2) a_n n(n-1) x^(n-2)
- sum(1) a_n n x^(n-1)
- sum(0) a_n x^(n)


sum(2) 2 a_n n(n-1) x^(n-1)
- sum(1) a_n n x^(n-1)
- sum(1) a_{n-1} x^(n-1)


- a_1 - a_0
+ sum(2) 2 a_n n(n-1) x^(n-1)
- sum(2) a_n n x^(n-1)
- sum(2) a_{n-1} x^(n-1)

- a_1 - a_0
+ sum(2) [2 a_n n(n-2)- a_{n-1}] x^(n-1)

zero out your coeffs

2 a_n n(n-2)- a_{n-1} = 0

a_n = a_{n-1}/(2n(n-2))

Answer 2

2xy'' = 0 + 2.2a2 x + 2.2.3a3 x^2 + 2.3.4a4 x^3+...
-y'= -a1 - 2a2 x - 3a3 x^2 -4a4 x^3 - ...
-y = -a0 - a1 x - a2 x^2 - a3 x^3 - ...

2.2a2 - 2a2 - a1 = 0
2a2(2-1) -a1 = 0
a2 = a1/(2(2-1)

2.2.3a3 -3a3 -a2 = 0
3a3(2.2-1) -a2 = 0
a3 = a2/(3(2.2-1))

2.3.4a^4 -4a4 -a3 = 0
4a4(2.3-1) -a3 = 0
a4 = a3/(4(2.3-1)

an = a_{n-1}/(n(2n-3))

did i mess up on this approach?

Answer 3

@amistre64 Notice that \(x=0\) is a singular point, so we can't find power series solution about \(x=0\). (You can see this by dividing through by \(x\):
\[y''-\frac{y'}{x}-\frac{y}{x}=0\]where \(x=0\) is a regular singular point; see http://mathworld.wolfram.com/RegularSingularPoint.html for more info.)

Answer 4

You have to make a slight modification to your approach. Instead of assuming a solution of the form
\[y=\sum_{n\ge0}a_nx^n\]you use
\[y=\sum_{n\ge0}a_nx^{n+k}\]where \(k\) is some real number - this is the standard idea behind the Frobenius method. Differentiating and substituting into the ODE yields
\[x\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}-\sum_{n\ge0}a_n(n+k)x^{n+k-1}-\sum_{n\ge0}a_nx^{n+k}=0\]
Distribute the \(x\) into the first series, and rewrite the third as
\[\sum_{n\ge0}a_nx^{n+k}=\sum_{n\ge1}a_{n-1}x^{n+k-1}\]so that we can group the series together by the \(x^{n+k-1}\) term. But first, we'll need to extract the first terms of the first and second series so that we can combine all three series. This way they'll start at the same index term, \(n=1\), and we have the compact form of the ODE,
\[a_0k(k-2)x^{k-1}+\sum_{n\ge1}\bigg(a_n(n+k)(n+k-2)-a_{n-1}\bigg)x^{n+k-1}=0\]

Answer 5

Now the same idea behind the usual power series approach can be used. Each power term should cancel, meaning the inidical equation (the term not belonging to the series above) can be solved for its roots and we can find an appropriate value for \(k\):
\[0=a_0k(k-2)x^{k-1}=0~~\implies~~k=0,k=2\]which gives us two (linearly independent) power series solutions in the form
\[y=\sum_{n\ge0}b_nx^n+\sum_{n\ge0}c_nx^{n+2}\]where \(b_n\) and \(c_n\) are solutions to the recurrence
\[b_nn(n-2)-b_{n-1}=0\quad\text{and}\quad c_nn(n+2)-c_{n-1}=0\]respectively.

Answer 6

(I have this slight suspicion I might have made a mistake somewhere, so take the final solution with a grain of salt...)

Answer 7

Yep, as I mentioned, the recurrence relations above generate the same solutions... The series with \(b_n\) is the same as the one with \(c_n\), just offset by 2 terms which happen to be zero... I think I'm mistaken about the two series solutions' forms, but everything up to that point seems correct. I'll look into this some more later.

Answer 8

when the roots of the indicial equation differ by an integer, you do not necessarily get two distinct solutions, as you noted

Answer 9

ohhhhhh boy... missed a negative sign in my calculation. Thanks for your help, after I reworked it I found my mistake.

Answer 10

I'll post my solution tomorrow. Thanks again for all of the help.

Answer 11

Ah okay, that makes sense. I was checking the solutions with Mathematica and it gave me a solution in terms of \(I_n\) and \(K_n\) (the modified Bessel functions of first/second kind, respectively), the first of which matched the \(b_n\) series, but I don't know enough about them say what the difference is on a technical level. Judging by the formulations on the Wiki page, I'm tempted to say they're not linearly independent, but I also suspect that's not true. I'd have to work it out for myself...

Answer 12

Here's my solution:
\[r_1=3/2\]

\[y_1=\sum_{n=0}^{\infty}a_nx^{n+\frac{3}{2}}\]
\[y_1'=\sum_{n=0}^{\infty}(n+\frac{3}{2})a_nx^{n+\frac{1}{2}}\]
\[y_1''=\sum_{n=0}^{\infty}(n+\frac{3}{2})(n+\frac{1}{2})a_nx^{n-\frac{1}{2}}\]

plugging into the ODE:

\[\sum_{n=0}^{\infty}2(n+\frac{3}{2})(n+\frac{1}{2})a_nx^{n+\frac{1}{2}}-\sum_{n=0}^{\infty}(n+\frac{3}{2})a_nx^{n+\frac{1}{2}}-\sum_{n=1}^{\infty}a_{n-1}x^{n+\frac{1}{2}}=0\]
now I change the index of the first 2 terms to get that \[a_0\in \mathbb{R}\] and then I get to the recurrence relation:
\[a_n=\frac{a_{n-1}}{n(2n+3)}\]
and then solve in to get that:
\[y_1=x^{\frac{3}{2}}(1+\sum_{n=1}^{\infty}\frac{3x^n}{n!(2n+3)!!})\]

Answer 13

Ah yeah you're right, I didn't notice the \(2\) in the first term. Your solution looks right to me.