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Thread: 2nd order linear differential equation

  1. #1
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    2nd order linear differential equation

    Can someone check my working for this question

    By making the substitution x=e^t, reduce the following linear differential equation to one with constant coefficients, and hence solve it:

    x^2\frac{d^2y}{dx^2}-4x\frac{dy}{dx}+6y=x^5 \ \ \ \ \ \ , x>0

    This is what I've done so far:

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

    \frac{dy}{dx}=\frac{dy}{dt}\cdot e^{-t}

    \frac{d^2y}{dx^2} = \frac{d}{dx}\left(e^{-t}\,\frac{dy}{dt}\right)

    = e^{-t}\,\frac{d}{dx}\left(\frac{dy}{dt}\right) + \frac{d}{dx}(e^{-t})\,\frac{dy}{dt}

    = e^{-t}\,\frac{d}{dt}\left(\frac{dy}{dt}\right)\,\frac{  dt}{dx} + \frac{d}{dt}(e^{-t})\,\frac{dt}{dx}\,\frac{dy}{dt}

    = e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}

    e^{2t}\left(e^{-2t}\,\frac{d^2y}{dt^2} - e^{-2t}\,\frac{dy}{dt}\right)-4e^t\left(e^{-t}\frac{dy}{dt}\right)+6y=e^{5t}

    \frac{d^2y}{dt^2}-5\frac{dy}{dt}+6y=e^{5t}

    Then would you solve that like a normal ODE
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  2. #2
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    Everything looks fine

    BTW thank you for making an effort to do the problem yourself and showing all your working
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