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Math Help - Differential equation change of variable

  1. #1
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    Differential equation change of variable

    I'm having some trouble getting to the solution of the following differential equation:

    Given that y=1 at x=2, use the substitution v=3x-y-3 to solve

    \left(3x-y-1\right)\frac{dy}{dx}=\left(3x-y+3\right)

    When I differentiate v with respect to x, I eventually find \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv} (my suspicion is that my error is in this step), and when I perform the relevant substitutions and integrate I do not get to the correct answer of

    2\ln|3x-y-3|=y-x+1+2\ln 2

    Any help on this would be greatly appreciated.
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  2. #2
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    Re: Differential equation change of variable

    Are we agree that

    v\ =\ 3x - y - 3\  \Rightarrow\ \frac{dv}{dx}\ =\ 3 - \frac{dy}{dx}

    and hence

    \frac{dy}{dx}\ =\ 3 - \frac{dv}{dx}

    ... ?

    That should give you a much easier ride than your

    \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}
    Thanks from britmath
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  3. #3
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    Re: Differential equation change of variable

    Quote Originally Posted by tom@ballooncalculus View Post
    Are we agree that

    ...and hence

    \frac{dy}{dx}\ =\ 3 - \frac{dv}{dx}

    That should give you a much easier ride than your

    \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}
    Yes, that's much easier! Thanks.
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