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Math Help - The second order ODE

  1. #1
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    Unhappy The second order ODE

    Help me with this

    y'' = 18\sin ^3y\cos y, y(1) =\frac{\pi}{2}, y'(1) = 3.

    I tried to solve it, but I got non-elementary integral
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  2. #2
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    Quote Originally Posted by Ment View Post
    Help me with this

    y'' = 18\sin ^3y\cos y, y(1) =\frac{\pi}{2}, y'(1) = 3.

    I tried to solve it, but I got non-elementary integral
    Note that \frac{d^2y}{dt^2} = f(y) \Rightarrow v \frac{dv}{dy} = f(y) where v = \frac{dy}{dt}.

    So solve for v (note that the DE is seperable) and then use your solution for v to solve for y.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Note that \frac{d^2y}{dt^2} = f(y) \Rightarrow v \frac{dv}{dy} = f(y) where v = \frac{dy}{dt}.

    So solve for v (note that the DE is seperable) and then use your solution for v to solve for y.
    Please give more tips.

    Here's my solution

    y'' = 18{\sin ^3}y\cos y, y(1) = \frac{\pi}{2},y'(1) = 3.

    y' = p(y) \Rightarrow y'' = p'(y)p(y)

    p'p = 18\sin^3y\cos y

    \int pdp  = 18\int \sin^3y\cos ydy

    \frac{p^2}{2} = \frac{9}{2}\sin^4y + C_1 \Rightarrow (y')^2 = 9\sin^4y + C_1

    y' = \pm \sqrt {9\sin^4y + C_1}

    x = \pm \int \frac{dy}{\sqrt{9\sin^4y + C_1}}

    I got non-elementary integral.
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  4. #4
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    I guess, then, the question is why you would expect to get a solution in terms of an elementary integral! That's a very non-linear equation and non-linear differential equations tend to have non-elementary solutions. Something as simple (comparatively) as y"= sin(y) has solutions that can only be written in terms of elliptic integrals.
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  5. #5
    Senior Member DeMath's Avatar
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    Quote Originally Posted by Ment View Post

    Here's my solution

    y'' = 18{\sin ^3}y\cos y, y(1) = \frac{\pi}{2},y'(1) = 3.

    y' = p(y) \Rightarrow y'' = p'(y)p(y)

    p'p = 18\sin^3y\cos y

    \int pdp  = 18\int \sin^3y\cos ydy

    \frac{p^2}{2} = \frac{9}{2}\sin^4y + C_1 \Rightarrow (y')^2 = 9\sin^4y + C_1

    y' = \pm \sqrt {9\sin^4y + C_1}

    x = \pm \int \frac{dy}{\sqrt{9\sin^4y + C_1}}

    I got non-elementary integral.
    You do not necessarily find a general solution to solve your problem of Cauchy.

    You have the initial conditions in the symmetrical form: x=1\ \leftrightarrow\ y={\pi\over2}\ \leftrightarrow\ y'=3
    Also you got \left(y'\right)^2 = 9\sin^4y + C_1. Now find constant C_1:

    \left\{\begin{gathered}\left(y'\right)^2 = 9\sin^4y + C_1, \hfill \\ y(1) = \frac{\pi}{2} ~\wedge~ y'\left( 1 \right) = 3; \hfill \\ \end{gathered}  \right. ~ \Rightarrow ~ 3^2 = 9 \sin^4\frac{\pi}{2} + C_1 ~ \Leftrightarrow ~ C_1 = 0.

    So you have

    \left(y'\right)^2 = 9\sin^4y ~ \Leftrightarrow ~ y' = 3\sin^2y ~ \Rightarrow ~ x = \frac{1}{3}\int \frac{dx}{\sin^2y}  = C_2 - \frac{1}{3}\cot y.

    Now find the constant C_2:

    \left\{ \begin{gathered}x = C_2 - \frac{1}{3}\cot y, \hfill \\ y\left( 1 \right) = \frac{\pi}{2}; \hfill \\ \end{gathered}  \right. ~ \Rightarrow ~ 1 = C_2 - \frac{1}{3}\cot\frac{\pi}{2} ~ \Leftrightarrow ~ C_2 = 1.


    Finally you have:

    x = 1 - \frac{1}{3}\cot y.
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