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Thread: 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 $\displaystyle y=1$ at $\displaystyle x=2$, use the substitution $\displaystyle v=3x-y-3$ to solve

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

    When I differentiate $\displaystyle v$ with respect to $\displaystyle x$, I eventually find $\displaystyle \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}$ (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

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

    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

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

    and hence

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

    ... ?

    That should give you a much easier ride than your

    $\displaystyle \frac{dy}{dx}=\tfrac{3}{2}\frac{dy}{dv}$
    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

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

    That should give you a much easier ride than your

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