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Math Help - separable differential equation dy/dx = y/x

  1. #1
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    separable differential equation dy/dx = y/x

    Question is to solve \frac{{dy}}{{dx}} = \frac{y}{x}

    I'm completely lost with these, this is as far as I get and then I'm not sure what to do. I get as far as this

    <br />
\begin{array}{l}<br />
  \equiv \int {\frac{{dy}}{y} = \int {\frac{{dx}}{x}} }  \\ <br />
  \equiv \int {\frac{1}{y}dy = \int {\frac{1}{x}dx} }  \\ <br />
 \end{array}<br />

    i know they need to be integrated from here so

    <br />
\ln y = \ln x + c<br />
    where c is my constant

    the answer given in the text is
    <br />
y = kx<br />
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  2. #2
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    Exponentiate both sides of the equation (raise the ln by e). Then you get y=x*e^(c). Because c is a constant and e is a number it can be absorbed and called k, thus you get y=kx
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  3. #3
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    thanks Jake. so with respect to the e^c , as it is just a number you are replacing is with another term to short note it? ie k=e^c
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  4. #4
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    yep
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  5. #5
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    Quote Originally Posted by Craka View Post
    Question is to solve \frac{{dy}}{{dx}} = \frac{y}{x}

    I'm completely lost with these, this is as far as I get and then I'm not sure what to do. I get as far as this

    <br />
\begin{array}{l}<br />
\equiv \int {\frac{{dy}}{y} = \int {\frac{{dx}}{x}} } \\ <br />
\equiv \int {\frac{1}{y}dy = \int {\frac{1}{x}dx} } \\ <br />
\end{array}<br />

    i know they need to be integrated from here so

    <br />
\ln {\color{red}|} y {\color{red}|} = \ln {\color{red}|} x {\color{red}|} + c<br />
    where c is my constant

    the answer given in the text is
    <br />
y = kx<br />
    The red modulus brackets which you neglected and which I've added are essential. Failure to include them means that your equation is wrong.
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  6. #6
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    Thanks I'd forgotten that.
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