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Nonlinear 2nd order ode reduction solutions

  1. Mar 30, 2012 #1
    hey guys

    i've been trying to work out this ode reduction question,

    http://img204.imageshack.us/img204/8198/asdawt.jpg [Broken]

    after i use the hint and end up with a seperable equation then integrate to get

    [itex]\begin{align}
    & p=\pm \frac{1}{\sqrt{{{y}^{2}}-2c}} \\
    & \text{Then}\,\,\,\text{integrating}\,\,\,\text{again (using}\,\,wolframalpha) \\
    & y=\pm \log \left( \sqrt{{{y}^{2}}-2c}+y \right)+c \\
    & A{{e}^{\pm y}}=\sqrt{{{y}^{2}}-2c}+y \\
    \end{align}[/itex]

    the first line above is consistent with what i get when i use mathematica

    but after getting p, is it correct to integrate p to get y? It doesn't look correct in my working but I can't find another way to do it.

    i thought about trying to solve the equation as y'= +/- ... instead as a first order ode but it seems too complicated,

    Is there something simple im not seeing?

    Thanks in advance
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 30, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, you are told that p= y'= dy/dx so
    [tex]\frac{dy}{dx}= \pm\frac{1}{\sqrt{y^2- 2c}}[/tex]
    and then
    [tex]\frac{dy}{\pm\sqrt{y^2- 2c}}= dx[/tex]

    Integrate both sides of that.
     
  4. Mar 30, 2012 #3
    Hey HallsofIvy

    Thanks for replying,

    Are you sure its not:
    [tex] \pm dy \sqrt{y^2- 2c}= dx[/tex]

    since you take the dx to the right side, and the inverted root onto the left?

    using the initial condition I get c=0 since

    [tex]\begin{align}
    & y'(0)=\pm \frac{1}{\sqrt{{{y}^{2}}(0)-2c}}=\pm \frac{1}{\sqrt{1-2c}}=-1 \\
    & \pm \frac{1}{\sqrt{1-2c}}=-1,\,\,\,\pm 1=-\sqrt{1-2c}\,\,squaring\,\,both\,\,sides \\
    & 1=1-2c \\
    & c=0 \\
    & So\,y'(x)=\pm \frac{1}{y} \\
    & \pm ydy=dx \\
    & \pm \frac{1}{2}{{y}^{2}}=x+c \\
    & \pm \frac{1}{2}{{y}^{2}}(0)=\pm \frac{1}{2}=c \\
    & \therefore y(x)=\sqrt{2x\pm \frac{1}{2}} \\
    \end{align}
    [/tex]

    does that look ok?
     
    Last edited: Mar 30, 2012
  5. Mar 31, 2012 #4
    I tried it again today with a fresh start and I ended up getting the same thing,
    The thing that makes me sus about it is the +/- inside the square root
     
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