Very basic Q about solns to y" = y

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
kostoglotov
Messages
231
Reaction score
6
Wolfram and the Linear Algebra text I'm currently working on, give the two possible solutions of [itex]\frac{d^2y}{dx^2}=y[/itex] as being [itex]e^{x}[/itex] and [itex]e^{-x}[/itex], or rather, constant multiples of them.

Here wolfram agrees:

http://www.wolframalpha.com/input/?i=d^2y/dx^2=y

My question is, why isn't [itex]y = e^{x} + x[/itex] and [itex]y = e^{-x} + x[/itex] not solutions?

Or is an added polynomial in the solution just part of some nullspace of special solutions that we consider separately in general?

edit: wouldn't the added polynomial change the dimension of the solution space?
 
Physics news on Phys.org
Why should y=e^x + x be a solution? y=x is not a solution.
 
davidmoore63@y said:
Why should y=e^x + x be a solution? y=x is not a solution.

If you take y=e^x + x and diff it twice...damn it...ok, I see it now...this was dumb.