EugP said:
Do you need to get a series solution? If not, if you substitue y'' = r', it seems it becomes a simple separation of variables problem.
Reduction of order won't work, as you have the second derivative
and the function.
To solve it with power series, you must take y(x)=\sum_0^\infty a_n x^n, so
y'(x)=\sum_{n=1}^\infty n a_n x^{n-1},
y''(x)=\sum_{n=2}^\infty n (n-1) a_n x^{n-2}.
If you plug it in the ode, you'll obtain the following equation:
\sum_{n=2}^\infty n (n-1) a_n x^{n-2}+\sum_{n=0}^\infty a_n x^{n+2}=0,
therefore, you'll have to change the index in the sums in order to have the same powers of x in both sums. To do that, take n=m+4 in the first sum. The equation becomes
\sum_{m=-2}^\infty (m+4) (m+3) a_{m+4} x^{m+2} + \sum_{n=0}^\infty a_n x^{n+2}=0.
Writing down the first two terms of the left sum and factorizing, we have
2 a_2+6 a_3 x + \sum_{n=0}^\infty [(n+4) (n+3) a_{n+4}+a_n]x^{n+2} =0.
The only way the equation is going to be fullfilled is that a_2=a_3=0, since there is no constants or first order powers under the sum sign. Hence,
\sum_{n=0}^\infty [(n+4) (n+3) a_{n+4}+a_n]x^{n+2} =0,
wich gives the recurrence relation
a_{n+4}=-\frac{a_n}{(n+4)(n+3)}.
Now, I've already done the hard part. You have to take it from here. To construct the first solution take a_0=1,a_1=0. To construct the second solution take a_0=0,a_1=1 (why?). From the relationship above, it's easy to see that the condition a_2=a_3=0, makes a hole bunch of terms to be zero, so you only have to calculate two general terms.
Good luck.
