By letting
[tex]y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}[/tex]
we get derivatives
[tex]y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}[/tex]
[tex]y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}[/tex]
a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to
[tex]5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0[/tex]
Examine the lowest degree term [tex]\left(x^{r-1}\right)[/tex], which gives rise to the indicial equation,
[tex]5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0[/tex]
with roots at r = 0 and r = 4/5.
b) The recurrence for the coefficients [tex]c_k[/tex] is
[tex](k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}[/tex]
so that with r = 4/5, the coefficients are governed by
[tex]c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}[/tex]
c) Starting with [tex]c_0=1[/tex], we find
[tex]c_1 = -\dfrac{c_0}5 = -\dfrac15[/tex]
[tex]c_2 = -\dfrac{c_1}{10} = \dfrac1{50}[/tex]
so that the first three terms of the solution are
[tex]\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}[/tex]
