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Math Help - Differential equation substitution

  1. #1
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    Differential equation substitution

    If anyone could explicitly show me the steps of how this substitution works I'd be v. grateful

    <br />
\frac{{dy}}<br />
{{dx}} = \frac{{x^2  + y^2  + 9xy}}<br />
{{9x^2 }} \Rightarrow \frac{{dy}}<br />
{{dx}} = \frac{1}<br />
{9}(1 + (\frac{y}<br />
{x})^2 ) + \frac{y}<br />
{x}<br />
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  2. #2
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    Now substitute y=xz\implies y'=z+xz' and replace this into the equation.
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  3. #3
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    is this substituion always valid for homogenous equations?
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  4. #4
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    As far as I know, yes.

    Unless someone wants to give another opinion.
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  5. #5
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    just quickly one more thing...

    Where does this v^2 come from in the last equation????

    <br />
\frac{{dy}}<br />
{{dx}} = \frac{{3y - x}}<br />
{{x + y}}<br />

    so use
    <br />
y = vx<br />

    <br />
v + xv' = \frac{{3v - 1}}<br />
{{1 + v}}<br />

    <br />
xv' = \frac{{3v - 1 - v - v^2 }}<br />
{{1 + v}}<br />
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  6. #6
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    \frac{3v-1}{1+v}-v=\frac{3v-1-v-v^{2}}{1+v}.

    It's just algebra.
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