Very basic Q about solns to y" = y

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1. Oct 6, 2015

kostoglotov

Wolfram and the Linear Algebra text I'm currently working on, give the two possible solutions of $\frac{d^2y}{dx^2}=y$ as being $e^{x}$ and $e^{-x}$, 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 $y = e^{x} + x$ and $y = e^{-x} + x$ 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?

2. Oct 6, 2015

davidmoore63@y

Why should y=e^x + x be a solution? y=x is not a solution.

3. Oct 6, 2015

kostoglotov

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