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I solved the following: \begin{align} \frac{dy}{y}&=\frac{dy'}{2y'+wy}\\ \ln y + \ln C &= \frac{1}{2} \ln \left(2y'+wy\right)\\ 2\ln y + \ln C &= \ln \left(2y'+wy\right)\\ C &= \frac{2y'}{y^2} + \frac{w}{y}\\ \end{align}

I isolated the constant since the purpose of solving is to find the differential invariant. However the answer given is: $$ \frac{y'}{y^2}+\frac{w}{y} $$

where am I going awry?

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    $\begingroup$ The DE is not separable so you can't integrate it the way you did it. $\endgroup$ – Aryadeva Jan 19 at 6:17
  • $\begingroup$ yes, I suspected I had oversimplified the situation in my haste. $\endgroup$ – sarah jamal Jan 19 at 12:51
  • $\begingroup$ Yes It happens sometimes. But the DE is not that complicated to integrate Sarah $\endgroup$ – Aryadeva Jan 19 at 13:05
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    $\begingroup$ you're right, thanks $\endgroup$ – sarah jamal Jan 19 at 14:14
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Your mistake is, $y \ \& \ y'$ aren't independent variables, so you can't suppose them as constant when you integrate with respect to the other ones! You must follow Exact Differential Equations method instead. By inspiration from given answer, Write it as follows, to see its exactness: $$\frac{dy}{y} = \frac{dy'}{2y'+\omega y} \Longrightarrow (2y'+\omega y)dy = ydy' \Longrightarrow ydy' - (2y'+\omega y)dy = 0 \Longrightarrow$$ $$\Longrightarrow \frac{1}{y^2}dy' + (\frac{-2y'}{y^3}+\frac{-\omega}{y^2})dy = 0 \Longrightarrow d(\frac{y'}{y^2}+\frac{\omega}{y}) = 0 \Longrightarrow \frac{y'}{y^2}+\frac{\omega}{y} = C$$

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$$\frac{dy}{y}=\frac{dy'}{2y'+wy}$$ $$(2y'+wy){dy}-y{dy'}=0$$ $$wy{dy}+2y'dy-y{dy'}=0$$ Multiply by $y$ $$wy^2{dy}+y'dy^2-y^2{dy'}=0$$ Divide by $y^4$: $$\dfrac w {y^2}{dy}+\dfrac {y'dy^2-y^2{dy'}}{y^4}=0$$ $$\dfrac w {y^2}{dy}-d\dfrac {y'}{y^2}=0$$ Integrate: $$-\dfrac w y- \dfrac {y'}{y^2}=k$$ $$\dfrac w y+ \dfrac {y'}{y^2}=C$$

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Using the rule $\frac ab=\frac cd\implies \frac ab=\frac cd=\frac{a+c}{b+d}$, add the $ω$ extended first term to the second to find $$ \frac{dy}{y}=\frac{dy'}{2y'+ωy}=\frac{dy'+ωdy}{2y'+2ωy} $$ Now taking the first and last term this can be directly integrated to $$ \ln|y|=\frac12\ln|y'+ωy|+c\implies Cy^2=y'+ωy $$

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