- #1
kostoglotov
- 234
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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?
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?