Question

how do you give a power series solution for y''-4y'-5y=0; y(0)=5, y'(0)=0?

Answer 1

please try this substitution:
\[\large y = c\exp \left( {\lambda x} \right)\]
where \[\lambda \] is a parameter, and c is a constant.
After that substitution, you should get a quadratic equation for \[\lambda \]

Answer 2

Looks like he wants a power series solution

Answer 3

yes! you are right! @dan815
another substitution can be this:

\[\large y = \sum\limits_{j = 0}^{ + \infty } {{a_j}{x^j}} \]

Answer 4

I know that way but the teacher was asking us to solve it using the power series method of
y=\[\sum_{n} a_{n}x ^{n}\]