<s>[INST] 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
 [/INST] $$\left( {{x^2} + xy + 4x + 2y + 4} \right)\frac{{dy}}{{dx}} - {y^2} = 0,$$
$$\left( {{{\left( {x + 2} \right)}^2} + y\left( {x + 2} \right)} \right)dy-y^2 dx=0$$
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 $$\frac{u+1}{u}du=-\frac{dt}{t}$$.
$$C+u+\ln{u}+\ln{t}=0$$
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