Question

Find values of m so that the function y xm is a solution of the given differential equation. x^2y''-7xy' 15y=0

Answer 1

If
`y=x^m` is a solution to the ODE

`x^2 y'' - 7xy' + 15y = 0`

then substituting
`y` and its derivatives

`y' = mx^(m-1) \text{ and } y'' = m(m-1)x^(m-2)`

gives

`m(m-1)x^2x^(m-2) - 7mxx^(m-1) + 15x^m = 0`

`m(m-1)x^(m) - 7mx^(m) + 15x^m = 0`

`(m(m-1) - 7m + 15)x^(m) = 0`

We ignore the trivial case of
`y=x^m=0`. Solve for
`m`.

`m(m-1) - 7m + 15 = 0`

`m^2 - 8m + 15 = 0`

`(m - 3) (m - 5) = 0`

`\implies \boxed{m=3} \text{ or } \boxed{m=5}`