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Math Help - Non Linear homogenous differential equation

  1. #1
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    Non Linear homogenous differential equation

    Just need a bit of help finishing this one, not sure if I'm going about this correctly...

    <br />
\begin{gathered}<br />
\frac{{dy}}<br />
{{dx}} = \frac{{y^2 + 3xy + x^2 }}<br />
{{x^2 }} \hfill \\<br />
\frac{{dy}}<br />
{{dx}} = \frac{{v^2 + 3v}}<br />
{x} - \frac{v}<br />
{x} \hfill \\<br />
\frac{{dy}}<br />
{{dx}} = v^2 + 2v/x \hfill \\<br />
\hfill \\<br />
\int {\frac{{dv}}<br />
{{v(v + 2v)}} = \int {\frac{{dx}}<br />
{x} = > \int {(\frac{1}<br />
{v} - \frac{1}<br />
{{1 + v}})dv = \int {\frac{{dx}}<br />
{x}} } } } \hfill \\<br />
\end{gathered} <br />


    <br />
\ln |v/2 + v| = \ln |x| + A<br />

    ......

    I need to know how to get to this....

    <br />
\frac{x}<br />
{{A - \ln |x|}} - x<br />
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  2. #2
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    Quote Originally Posted by dankelly07 View Post
    Just need a bit of help finishing this one, not sure if I'm going about this correctly...

    <br />
\begin{gathered}<br />
\frac{{dy}}<br />
{{dx}} = \frac{{y^2 + 3xy + x^2 }}<br />
{{x^2 }} \hfill \\<br />
\frac{{dy}}<br />
{{dx}} = \frac{{v^2 + 3v}}<br />
{x} - \frac{v}<br />
{x} \hfill \\<br />
\frac{{dy}}<br />
{{dx}} = v^2 + 2v/x \hfill \\<br />
\hfill \\<br />
\int {\frac{{dv}}<br />
{{v(v + 2v)}} = \int {\frac{{dx}}<br />
{x} = > \int {(\frac{1}<br />
{v} - \frac{1}<br />
{{1 + v}})dv = \int {\frac{{dx}}<br />
{x}} } } } \hfill \\<br />
\end{gathered} <br />


    <br />
\ln |v/2 + v| = \ln |x| + A<br />

    ......

    I need to know how to get to this....

    <br />
\frac{x}<br />
{{A - \ln |x|}} - x<br />
      \frac{dy}{dx} = \frac{y^2 + 3xy + x^2 }{x^2}

    With y = vx substitution:

     v + x\frac{dv}{dx} = v^2 + 3v + 1
     x\frac{dv}{dx} = v^2 + 2v + 1 = (v+1)^2
    \int \frac{dv}{(v+1)^2} = \int \frac{dx}{x}
     -\frac1{v+1} = \ln C|x|
     v = -\frac1{\ln C|x|} - 1 \Rightarrow y = -\frac{x}{\ln C|x|} - x

    So the solution is y = \frac{x}{A - \ln|x|} - x where A = - \ln C, where A is as arbitrary as C is.*


    *- These are Mr.Fantastic's words... I like it.
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