General solution of Differential equation What is solution of the differential equation:
\begin{aligned}x({yy'' + y'^2}) + yy' = 0\end{aligned}
What i am confused about it is the treatment of
$$\left(\dfrac{dy}{dx}\right)^2$$
for solving this question. The given answer is \begin{aligned}ax^2 + by^2 = C\end{aligned}
 A: A start: Let $u=yy'$. Then $x\frac{du}{dx}+u=0$. 
I expect you can finish from here.
A: Note that:
Given $$y^2$$
If we take the first derivative by implicit differentiation we get
$$2yy'$$
Again taking a derivative we get 
$$2(y')^2 + 2y''$$
This leads us to guess $u = y^2$ is a wise substitution.
From it the equation becomes
$$\frac{1}{2}(xu'') + \frac{1}{2}u' = 0$$
Now we divide out the 1/2 and make another substitution $u' = w$ to obtain:
$$xw' + w = 0$$
You can take it from here, once solved go backwards through the chain of substitutions 
A: I took a different approach and got the same answer as those given above.  Since
$$
x(yy''+y'^2)+yy'=0
$$is equivalent to
$$
x\frac{d}{dx}(yy')=-yy'
$$or
$$
\frac{\frac{d}{dx}(yy')}{yy'}=-\frac{1}{x}
$$then
$$
\frac{d}{dx}ln(yy')=-\frac{1}{x}
$$This integrates to give
$$
ln(yy')=ln(x^{-1})+lnC_1
$$which simplifies to
$$
yy'=\frac{C_1}{x}
$$From here I used a Lie group $G(x,y)=(\lambda x,\lambda^\beta y), \lambda_o=1$ with stabilizers $\mu=\frac{y}{x^\beta}$ and $\nu=\frac{y'}{x^{\beta -1}}$ to rewrite the equation.
$$
\lambda^\beta y \lambda^{\beta -1} y'=\frac{C_1}{\lambda x}
$$For the $\lambda$ terms to cancel out, $\beta =0$.  Therefore, the first two nontrivial group stabilizers are $\mu=y$ and $\nu=xy'$.  Rewriting the ODE in terms of its stabilizers, 
$$
\nu=\frac{C_1}{\mu}
$$Since
$$
x\frac{d\mu}{dx}=\nu=\frac{C_1}{\mu}
$$we have
$$
\frac{\mu ^2}{2}=C_1ln(x)+C_2 
$$Allowing the constants to absorb the factor of two and remembering that $\mu=y$,
$$
y=\sqrt{C_1ln(x)+C_2}
$$There's nothing original about this solution, but I thought you might enjoy seeing an alternate approach to the problem itself.  I picked up the Lie theory in Oak Ridge, and since so few people have ever seen it I like to share it when I can.  
