Question

Please help me with the below question.

Answer 1

By letting

`y = \displaystyle \sum_(n=0)^\infty c_n x^(n+r)`

we get derivatives

`y' = \displaystyle \sum_(n=0)^\infty (n+r) c_n x^(n+r-1)`

`y'' = \displaystyle \sum_(n=0)^\infty (n+r) (n+r-1) c_n x^(n+r-2)`

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

`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`

Examine the lowest degree term
`\left(x^(r-1)\right)`, which gives rise to the indicial equation,

`5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0`

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients
`c_k` is

`(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_(k+1) = 0 \implies c_(k+1) = -(c_k)/(5k+5r+1)`

so that with r = 4/5, the coefficients are governed by

`c_(k+1) = -(c_k)/(5k+5) \implies \boxed{g(k) = -\frac1{5k+5}}`

c) Starting with
`c_0=1`, we find

`c_1 = -\frac{c_0}5 = -\frac15`

`c_2 = -(c_1)/(10) = \frac1{50}`

so that the first three terms of the solution are

`\displaystyle \sum_(n=0)^2 c_n x^(n + 4/5) = \boxed{x^(4/5) - \frac15 x^(9/5) + \frac1{50} x^(13/5)}`