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Thread: 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...

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


    $\displaystyle
    \ln |v/2 + v| = \ln |x| + A
    $

    ......

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

    $\displaystyle
    \frac{x}
    {{A - \ln |x|}} - x
    $
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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...

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


    $\displaystyle
    \ln |v/2 + v| = \ln |x| + A
    $

    ......

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

    $\displaystyle
    \frac{x}
    {{A - \ln |x|}} - x
    $
    $\displaystyle \frac{dy}{dx} = \frac{y^2 + 3xy + x^2 }{x^2}$

    With $\displaystyle y = vx$ substitution:

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

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


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