Solve the diffrential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$ A solution curve of the differential equation $\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0$, $x>0$ passes through the point (1,3). Find the solution curve.
I am not able to proceed as I am not able to convert the standard differentiable form
 A: $$\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0,$$
$$\left( {{{\left( {x + 2} \right)}^2} + y\left( {x + 2} \right)} \right)dy-y^2 dx=0$$
After substituting $x+2$ with $t$, we get
$$ (t^2+yt)dy-y^2dt=0$$
After multiplying with $t^{-2}$, we get
$$(1+\frac{y}{t})\frac{dy}{dt}-(\frac{y}{t})^2=0$$
Now use substitution $y=tu$, $y'=u+tu'$ from where is $(1+u)(u+tu')-u^2=0$
and
$$u+tu'=\frac{u^2}{u+1}$$
From here we get $$tu'=-\frac{u}{u+1}$$
and finally $$\frac{u+1}{u}du=-\frac{dt}{t}$$.
$$C+u+\ln{u}+\ln{t}=0$$
A: Starting from @alans'answer
$$\left(1+\frac{y}{t}\right)y'-\left(\frac{y}{t}\right)^2=0$$ Using $y=t u$, this makes
$$(u+1) \left(t u'+u\right)-u^2=0$$ Switching variables
$$(u+1) \left(\frac t{t'}+u\right)-u^2=0\implies \frac {t'}t=-\frac{1}{u}-1$$
$$\log(t)+c=-u-\log (u)\implies k t=\frac{e^{-u}}{u}$$ This is the implicit solution.
If you want to go further, as already commented, you must use Lambert function; this would give
$$u=W\left(\frac{1}{k t}\right)$$ Back to $x$ and $y$
$$y=(x+2)\,W\left(\frac{1}{k (x+2) }\right)$$ Using the condition
$$3=3 \,W\left(\frac{1}{3k }\right)\implies k=\frac{1}{3 e}$$ SO, finally
$$y=(x+2)\,W\left(\frac{3e}{ x+2 }\right)$$
A: As suggested, you can use:
$$\left((x+2)^2+y(x+2)\right)\frac{dy}{dx}-y^2=0$$
if we first try with $$X=x+2,dX=dx:$$
$$(X^2+yX)\frac{dy}{dX}-y^2=0$$
$$X(X+y)\frac{dy}{dX}-y^2=0$$
this term of $X$ and $y$ together makes it quite hard to solve by hand
