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Math Help - Differential Equation with Substitution

  1. #1
    Super Member craig's Avatar
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    Differential Equation with Substitution

    Hi, hoping that someone here can help shed some light on this for me.

    You are given that x = e^t.

    Show that:

    a) x\frac{dy}{dx} = \frac{dy}{dt} - this was easy enough, could do this.

    b) x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}

    Here's how far I got:

    Using part a

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

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

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

    I have no idea what to do with the last, x\frac{d}{dt}(\frac{dy}{dx}) part, any help is greatly appreciated

    Thanks in advance
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  2. #2
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    Quote Originally Posted by craig View Post
    Hi, hoping that someone here can help shed some light on this for me.

    You are given that x = e^t.

    Show that:

    a) x\frac{dy}{dx} = \frac{dy}{dt} - this was easy enough, could do this.

    b) x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}

    Here's how far I got:

    Using part a

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

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

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

    I have no idea what to do with the last, x\frac{d}{dt}(\frac{dy}{dx}) part, any help is greatly appreciated

    Thanks in advance
    If \frac{dy}{dt} = x\frac{dy}{dx}, then wouldn't \frac{dy}{dx} = x^{-1}\frac{dy}{dt}?

    Also remember that x = e^t

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


    So wouldn't x \frac{d}{dt}\left(\frac{dy}{dx}\right) = e^t \frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right).


    Use the product rule to evaluate...
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  3. #3
    Super Member craig's Avatar
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    Quote Originally Posted by Prove It View Post
    If \frac{dy}{dt} = x\frac{dy}{dx}, then wouldn't \frac{dy}{dx} = x^{-1}\frac{dy}{dt}?

    Also remember that x = e^t

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


    So wouldn't x \frac{d}{dt}\left(\frac{dy}{dx}\right) = e^t \frac{d}{dt}\left(e^{-t}\frac{dy}{dt}\right).


    Use the product rule to evaluate...
    Thanks for the reply.

    I worked out x \frac{d}{dt}\left(\frac{dy}{dx}\right) to be e^t\left(-e^{-t}\frac{dy}{dt} + \frac{d^2y}{dt^2}e^{-t}\right) which simplifies to be \frac{d^2y}{dt^2} - \frac{dy}{dt}?

    Putting this back into the earlier equation I get:

    \frac{d^2y}{dt^2} = \frac{dy}{dt} + \frac{d^2y}{dt^2} - \frac{dy}{dt}

    Which simplifies again to be 0 = 0.
    ...
    Hmm, I seem to have 1. Proved the obvious that 0 = 0, and 2. Made a mistake somewhere along the line.

    Any ideas?

    Thanks again
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  4. #4
    Super Member craig's Avatar
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    Just for anyone reading through this at any point, here's the answer I have finally worked out!

    x\frac{d}{dt}(\frac{dy}{dx}) can be written as x\frac{d}{dx}(\frac{dy}{dx})\frac{dx}{dt}

    Which gives us x(\frac{d^2y}{dx^2})\frac{dx}{dt}

    We know from the given substitution that x = e^t, therefore \frac{dx}{dt} = e^t = x

    We then get x(\frac{d^2y}{dx^2})x

    Putting this last bit back into the equation, you have proved that x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}

    Took me 2 days but have finally found a way to do it

    *runs off in joy*
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