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Why can I not do this?

  1. Jun 26, 2015 #1
    If I have a DE like this:
    [tex]y''=\frac{3}{y^{2}}+5[/tex]
    Why can't I just integrate both sides to get:
    [tex]y'=-\frac{3}{y^{1}}+5y[/tex]?
    And then integrate again to solve for y?
     
  2. jcsd
  3. Jun 26, 2015 #2

    PeroK

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    Because ##y''## means ##\frac{d^2y}{dt^2}##

    So, you would need to integrate both sides wrt ##t##.

    If you let ##\frac{d^2f}{dy^2} = \frac{3}{y^{2}}+5##

    Where ##f## is a function of ##y##, then you can simply integrate both sides wrt ##y##.
     
  4. Jun 26, 2015 #3
    Oh that was really dumb of me, didn't think about the notation enough. I'm not supposed to actually solve this, I was just wondering. It looks like the solution is extremely complicated so I probably can't anyway: http://www.wolframalpha.com/input/?i=y''=3/y^2+5

    In fact, the solution for just y''=1/y looks complicated: http://www.wolframalpha.com/input/?i=y''=1/y

    I haven't actually studied non-linear DEs yet.
     
  5. Jun 26, 2015 #4

    pasmith

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    ODEs of the form [itex]y''(x) = f(y(x))[/itex] can in principle be solved by multiplying by [itex]y'[/itex] and integrating with respect to [itex]x[/itex] to obtain [tex]
    \frac12(y')^2 = F(y)[/tex] where [itex]F(y) = \frac12 y'(0)^2 + \int_{y(0)}^{y} f(s)\,ds[/itex] is an antiderivative of [itex]f[/itex]. The resulting first-order ODE is separable: [tex]
    y' = \pm\sqrt{2F(y)}
    [/tex] where some care is needed in determining the sign.
     
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