Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Very basic Q about solns to y" = y

  1. Oct 6, 2015 #1
    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?
     
  2. jcsd
  3. Oct 6, 2015 #2
    Why should y=e^x + x be a solution? y=x is not a solution.
     
  4. Oct 6, 2015 #3
    If you take y=e^x + x and diff it twice...damn it...ok, I see it now...this was dumb.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Very basic Q about solns to y" = y
  1. Y'' = -y (Replies: 4)

  2. Opinion about y =f(y) (Replies: 12)

  3. Y*dy/dx= y*Q(x)+ R(x) (Replies: 2)

Loading...