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

Homework Help: Initial Value Problem

  1. Jul 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Solve the initial value problem for y.

    [tex]\frac{d^2y}{dx^2}\,=\,\frac{1}{a}\sqrt{1\,+\,\left(\frac{dy}{dx}\right)^2}[/tex]

    [tex]y(0)\,=\,a[/tex] & [tex]y\prime(0)\,=\,0[/tex]

    a is a non-zero constant.



    2. Relevant equations

    The calculus.



    3. The attempt at a solution

    [tex]y\prime\,=\,sinh\,x[/tex] [tex]\longrightarrow[/tex] [tex]dy\prime\,=\,cosh\,x\,dx[/tex]

    [tex]1\,+\,sinh^2\,x\,=\,cosh^2\,x[/tex]

    [tex]dy\prime\,=\,\frac{1}{a}\,\sqrt{cosh^2\,x}\,dx\,=\,\frac{cosh\,x}{a}\,dx[/tex]

    [tex]\int dy\prime\,=\,\frac{1}{a}\,\int cosh\,x\,dx[/tex]

    [tex]y\prime\,=\,\frac{1}{a}\,sinh\,x\,+\,C[/tex]

    C = 0 from the initial condition [itex]y\prime(0)\,=\,0[/itex]

    [tex]\int y\prime\,=\,\frac{1}{a}\,\int sinh\,x\,dx[/tex]

    [tex]y\,=\,\frac{1}{a}\,cosh\,x\,+\,C[/tex]

    [tex]\frac{1}{a}\,(1)\,+\,C\,=\,a[/tex]

    [tex]C\,=\,a\,-\,\frac{1}{a}[/tex]

    [tex]y\,=\,\frac{1}{a}\left(cosh\,x\,+\,a^2\,-\,1\right)[/tex]

    That doesn’t match the answer that maple gives, what did I do wrong?
     
    Last edited: Jul 12, 2007
  2. jcsd
  3. Jul 13, 2007 #2

    If
    [tex]y\prime\,=\,\frac{1}{a}\,sinh\,x[/tex]
    then
    [tex]dy\prime\,=\,\sqrt{1+y\prime^2}\, dx \,=\,\sqrt{1+\frac{(sinh\,x)^2}{a^2}}\, dx \,\neq\,\frac{1}{a}\,cosh\,x\,dx[/tex]
     
    Last edited: Jul 13, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook