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Math Help - problem with differentiation

  1. #1
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    problem with differentiation

    if x = √ 1 - t^2 , transform the differential equation

    x (1 - x^2) d^2y/dx^2 + (1 - 3x^2) dy/dx - xy = 0

    the answer to this problem is below:-

    t (1 - t^2) d^2y/dt^2 + (1 – 3t^2) dy/dt - ty = 0



    question is that how come the answer is similar to the question.
    i dont think i just need to replace the x by t..
    im confused..is something wrong with the answer??
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  2. #2
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    Quote Originally Posted by slash View Post
    if x = √ 1 - t^2 , transform the differential equation

    x (1 - x^2) d^2y/dx^2 + (1 - 3x^2) dy/dx - xy = 0

    the answer to this problem is below:-

    t (1 - t^2) d^2y/dt^2 + (1 – 3t^2) dy/dt - ty = 0
    Chain rule here.

    x(1-x^2) \frac{d^2 y}{dx^2} + (1-3x^2) \frac{dy}{dx} - xy = 0

    Note, \frac{dy}{dx} = \frac{dy}{dt}\cdot \frac{dt}{dx}.
    Und, \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d\left( \frac{dy}{dx} \right)}{dt} \cdot \frac{dt}{dx} = \frac{d\left( \frac{dy}{dt}\frac{dt}{dx} \right)}{dt} \cdot \frac{dt}{dx} = \frac{d^2 y}{dt^2}\cdot \left( \frac{dt}{dx}\right)^2
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