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Differential equations

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    Solve [tex] \frac{d^2 y}{dt^2} = y [/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex] \frac{dy}{dt} = v [/tex]

    [tex] \frac{d^2 y}{dt^2} = v \frac{dv}{dy} [/tex]


    [tex] v \frac{dv}{dy} = y [/tex]

    [tex] v dv = y dy [/tex]

    [tex] \int v dv = \int y dy [/tex]

    [tex] v^2 = y^2 + C [/tex]

    [tex] ( \frac{dy}{dt} )^2 = y^2 + C [/tex]

    [tex] \frac{dy}{dt} = \sqrt{ y^2 + C } [/tex]

    [tex] \int \frac{dy}{ \sqrt{ y^2 + C }} = \int dt [/tex]

    [tex] \int \frac{dy}{ \sqrt{ y^2 + C }} = t [/tex]

    I have no idea how to integrate this. Have I gone wrong somewhere?

    Thanks in advance for any help.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Nov 19, 2008
  2. jcsd
  3. Nov 19, 2008 #2

    Mark44

    Staff: Mentor

    There is one function whose derivative is exactly equal to the function itself. It follows that the second derivative would also be equal to the original function.
     
  4. Nov 19, 2008 #3
    Oh wow how did I not see that.

    Thanks for the help!!
     
  5. Nov 19, 2008 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The is a "linear equation with constant coefficients" so there is an easy solution as Mark44 pointed out.

    But your method is interesting- and right! To integrate
    [tex]\int \frac{dy}{\sqrt{y^2+1}}[/tex]
    use a "hyperbolic" substitution. Let y= sinh(u). Since sinh2(u)+ 1= cosh2(u), [itex]\sqrt{y^2+ 1}= \sqrt{cosh^2(u)}= cosh(u)[/itex]. Also du= cosh(u) du so this becomes simply
    [tex]\int du= u+ C2= sinh(y)+ C[/tex]

    Of course, since
    [tex]sinh(y)= \frac{e^y- e^{-y}}{2}[/tex]
    You can see how that fits the "usual" solution.
     
  6. Nov 19, 2008 #5
    I see, quite true. I don't know too much about hyperbolic functions but I see what your saying. Thanks for the help!
     
  7. Nov 22, 2008 #6
    But should

    [tex] u = sinh^-^1 (y) [/tex] ?
     
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