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Thread: Diff EQ using separation of variables

  1. #1
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    Diff EQ using separation of variables

    I am stumped on solving the following diff eq:

    $\displaystyle \frac{\,dy}{\,dx}=\frac{y}{3+t}$ with the initial condition $\displaystyle y(0)=1$

    I have figured the general solution to be $\displaystyle y=3+t+C$,
    from $\displaystyle \ln\left|y\right|=\ln\left|3+t\right|+C$,
    but it doesn't yield the correct result: $\displaystyle y=\frac{1}{3}(3+t)$
    I keep getting $\displaystyle y=1+t$.

    Please, what am I doing wrong?

    Thank you,
    Jen
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  2. #2
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    I'll assume the $\displaystyle \frac{dy}{dx}$ in your original post was a typo ...

    $\displaystyle
    \frac{dy}{dt} = \frac{y}{3+t}
    $

    $\displaystyle \frac{dy}{y} = \frac{dt}{3+t}$

    $\displaystyle
    \ln|y| = \ln|3+t| + C
    $

    $\displaystyle y(0) = 1$ ...

    $\displaystyle \ln(1) = \ln(3) + C$

    $\displaystyle
    C = -\ln(3)
    $

    $\displaystyle
    \ln|y| = \ln|3+t| - \ln(3)
    $

    $\displaystyle \ln|y| = \ln\left|\frac{3+t}{3}\right|$

    $\displaystyle
    y = \frac{3+t}{3} = \frac{1}{3}(3 + t)
    $
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  3. #3
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    Thanks!

    Yes, I am sorry...that WAS a typo..

    Thank you so much!!


    Quote Originally Posted by skeeter View Post
    I'll assume the $\displaystyle \frac{dy}{dx}$ in your original post was a typo ...

    $\displaystyle
    \frac{dy}{dt} = \frac{y}{3+t}
    $

    $\displaystyle \frac{dy}{y} = \frac{dt}{3+t}$

    $\displaystyle
    \ln|y| = \ln|3+t| + C
    $

    $\displaystyle y(0) = 1$ ...

    $\displaystyle \ln(1) = \ln(3) + C$

    $\displaystyle
    C = -\ln(3)
    $

    $\displaystyle
    \ln|y| = \ln|3+t| - \ln(3)
    $

    $\displaystyle \ln|y| = \ln\left|\frac{3+t}{3}\right|$

    $\displaystyle
    y = \frac{3+t}{3} = \frac{1}{3}(3 + t)
    $
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