Solving a PDE by method of characteristics 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?
 A: 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:
$$\frac{dy}{y} = \frac{dy'}{2y'+\omega y} \Longrightarrow (2y'+\omega y)dy = ydy' \Longrightarrow ydy' - (2y'+\omega y)dy = 0 \Longrightarrow$$
$$\Longrightarrow \frac{1}{y^2}dy' + (\frac{-2y'}{y^3}+\frac{-\omega}{y^2})dy = 0 \Longrightarrow d(\frac{y'}{y^2}+\frac{\omega}{y}) = 0 \Longrightarrow \frac{y'}{y^2}+\frac{\omega}{y} = C$$
A: $$\frac{dy}{y}=\frac{dy'}{2y'+wy}$$
$$(2y'+wy){dy}-y{dy'}=0$$
$$wy{dy}+2y'dy-y{dy'}=0$$
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:
$$-\dfrac w y- \dfrac {y'}{y^2}=k$$
$$\dfrac w y+ \dfrac {y'}{y^2}=C$$
A: 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
$$
\ln|y|=\frac12\ln|y'+ωy|+c\implies Cy^2=y'+ωy
$$
