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Math Help - 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:

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

    I have figured the general solution to be y=3+t+C,
    from \ln\left|y\right|=\ln\left|3+t\right|+C,
    but it doesn't yield the correct result: y=\frac{1}{3}(3+t)
    I keep getting 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 \frac{dy}{dx} in your original post was a typo ...

     <br />
\frac{dy}{dt} = \frac{y}{3+t}<br />

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

     <br />
\ln|y| = \ln|3+t| + C<br />

    y(0) = 1 ...

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

     <br />
C = -\ln(3)<br />

     <br />
\ln|y| = \ln|3+t| - \ln(3)<br />

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

     <br />
y = \frac{3+t}{3} = \frac{1}{3}(3 + t)<br />
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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 \frac{dy}{dx} in your original post was a typo ...

     <br />
\frac{dy}{dt} = \frac{y}{3+t}<br />

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

     <br />
\ln|y| = \ln|3+t| + C<br />

    y(0) = 1 ...

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

     <br />
C = -\ln(3)<br />

     <br />
\ln|y| = \ln|3+t| - \ln(3)<br />

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

     <br />
y = \frac{3+t}{3} = \frac{1}{3}(3 + t)<br />
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