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Math Help - Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE HELP]

  1. #1
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    Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE HELP]

    I'm given that an Euler equation (*) is of the form:
    Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.15.53-pm.png
    where a and b are constants. Then I'm told, let x=lnt, and calculate dy/dt and d2y/dt2 in terms of dy/dx and d2y/dx2.

    I found:
    Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.19.52-pm.png
    But then I had problems with d2y/dt2.

    This is what I got:
    Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.23.08-pm.png
    Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.26.10-pm.png
    then subbing into (*) is good, except for d2y/dtdx term.


    Any help would be greatly appreciated
    Attached Thumbnails Attached Thumbnails Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.07.55-pm.png   Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0   [CHAIN RULE HELP]-screen-shot-2014-03-07-6.27.24-pm.png  
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  2. #2
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    Re: Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE H

    $\dfrac{d^2y}{dt^2}=\dfrac{d}{dt}\left(\dfrac{1}{t } \dfrac{dy}{dx}\right)=$

    $-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dt}\left(\dfrac{dy}{dx}\right)=$

    $-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dx}\left(\dfrac{dy}{dt}\right)=$

    $-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t}\dfrac{d} {dx}\left(\dfrac{dy}{dx}\dfrac{dx}{dt}\right)=$

    $-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t^2}\dfrac{ d}{dx}\left(\dfrac{dy}{dx}\right)=$

    $-\dfrac{1}{t^2}\dfrac{dy}{dx}+\dfrac{1}{t^2}\dfrac{ d^2y}{dx^2}=$

    $\dfrac{1}{t^2}\left(\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}\right)$
    Thanks from mathlover700
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    Re: Deriving Euler's Equation: from: t2(d2y/dt2) + at(dy/dt) + by = 0 [CHAIN RULE H

    Thank you very much for you help! I understand it now!
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