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Thread: differential equations help second order non homo

  1. #1
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    differential equations help second order non homo

    differential equations help second order non homo-asdd.png
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  2. #2
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    Re: differential equations help second order non homo

    Have you tried what they have suggested?
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    Re: differential equations help second order non homo

    yes but the lecturer just copied the correction from a page and skip a lot of step,so it is not making sense at all and
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    Re: differential equations help second order non homo

    Quote Originally Posted by Musawwir View Post
    yes but the lecturer just copied the correction from a page and skip a lot of step,so it is not making sense at all and
    Follow the instructions

    a) $x=e^t,~z(t)=y(x(t))$

    I will use upper dot notation to denote time derivatives and prime notation to denote derivatives of $y$ with respect to $x$ I.e.

    $\dot{z} = \dfrac{dz}{dt}$

    $y^\prime = \dfrac{dy}{dx}$

    $\dot{z} = y^\prime \dot{x} = y^\prime e^t$

    $y^\prime = \dot{z} e^{-t}$

    $\begin{align*}
    &\ddot{z}= y^{\prime\prime}(\dot{x}^2) + y^\prime \dot{x} \\
    &=y^{\prime\prime} e^{2t} + y^\prime e^t \\
    &=y^{\prime\prime} e^{2t} + \dot{z}
    \end{align*}$

    $y^{\prime\prime} = (\ddot{z}-\dot{z})e^{-2t}$

    Now you plug $y^\prime, ~y^{\prime\prime}$ into the original differential equation, letting $x=e^t$, and solve it in $t$

    Finally let $t = \ln(x)$
    Last edited by romsek; Dec 2nd 2017 at 11:50 AM.
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    Re: differential equations help second order non homo

    this is the only line i was not understanding zĘ=y′′(x˙2)+y′x
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    Re: differential equations help second order non homo

    Quote Originally Posted by Musawwir View Post
    this is the only line i was not understanding zĘ=y′′(x˙2)+y′x
    $\dot{z} = y^\prime \dot{x}$

    $\begin{align*}

    &\dfrac{d}{dt} \dot{z} = \\

    &\left(\dfrac{d}{dt} y^\prime\right)\dot{x} + y^\prime \ddot{x} = \\

    &\left(\dfrac{d}{dx} y^\prime \dot{x}\right)\dot{x} + y^\prime \ddot{x} = \\

    &\left(y^{\prime\prime}\dot{x}\right)\dot{x} + y^\prime \ddot{x} = \\

    &y^{\prime\prime}\dot{x}^2 + y^\prime \ddot{x}

    \end{align*}$
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