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Math Help - Second linear differential euqation with variable coefficient how to solve

  1. #1
    MHF Contributor Amer's Avatar
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    Second linear differential euqation with variable coefficient how to solve

    (1-x^2)y'' - xy' +4y = 2x (1 -x^2)^1/2

    Hint use the substitution x =sin t
    I used it and end with

    cos t y'' + sin t y' - (sin t)/(cos t) y' + 4y = 2 sin t (cos t)

    how to solve this i just want the name of the method
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  2. #2
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    Re: Second linear differential euqation with variable coefficient how to solve

    Hi !
    There is a mistake in your substitution.
    Show the details of what you have done and it will be possible to see where is the snag.
    Thanks from Amer
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  3. #3
    Behold, the power of SARDINES!
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    Re: Second linear differential euqation with variable coefficient how to solve

    Quote Originally Posted by Amer View Post
    (1-x^2)y'' - xy' +4y = 2x (1 -x^2)^1/2

    Hint use the substitution x =sin t
    I used it and end with

    cos t y'' + sin t y' - (sin t)/(cos t) y' + 4y = 2 sin t (cos t)

    how to solve this i just want the name of the method
    If looks like you just made an error in your dervaitves. You are really close!

    Suppose that y is a function of t

    y(t) and x=\sin(t)

    Then by the chain rule we have

    \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}

    We can take the derivative again using the product and chain rule to get

    \frac{d^2y}{dx^2} =\left( \frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}+\frac{dy}{  dx}\frac{d^2x}{dt^2}

    If you work the above derivatives you will get the first two terms of your ODE.

    Best wishes.

    TES
    Thanks from Amer
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  4. #4
    MHF Contributor Amer's Avatar
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    Re: Second linear differential euqation with variable coefficient how to solve

    Quote Originally Posted by TheEmptySet View Post
    If looks like you just made an error in your dervaitves. You are really close!

    Suppose that y is a function of t

    y(t) and x=\sin(t)

    Then by the chain rule we have

    \frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}

    We can take the derivative again using the product and chain rule to get

    \frac{d^2y}{dx^2} =\left( \frac{dx}{dt}\right)^2\frac{d^2y}{dx^2}+\frac{dy}{  dx}\frac{d^2x}{dt^2}

    If you work the above derivatives you will get the first two terms of your ODE.

    Best wishes.

    TES
    My question is
    (1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + 4y = 2x \sqrt{1-x^2}

    The sub is
     x = \sin t \Rightarrow \dfrac{dx}{dt} = \cos t \Rightarrow \dfrac{d^2x}{dt^2} = -\sin t

    \dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt}

    I made a mistake in follwing, i write it like below but when i differentiate with respect to t,
    \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}
    the left hand side I write it like this
    \dfrac{d^2y}{dx^2} which should be like this \dfrac{d^2y}{dx^2} \dfrac{dy}{dt}

    Thanks very much both
    Last edited by Amer; October 9th 2012 at 09:48 PM.
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  5. #5
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    Re: Second linear differential euqation with variable coefficient how to solve

    Quote Originally Posted by Amer View Post
    \dfrac{d^2y}{dx^2} which should be like this \dfrac{d^2y}{dx^2} \dfrac{dy}{dt}
    No, that is your mistake.
    In order to obtain dy/dx you have to write :
    dy/dx = dz/dx = (dz/dt)/(dx/dt) where z = dy/dx
    Frist, express dy/dx as an expression which contains only (dy/dt) and t , without any x in it.
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