How do I find the singular solution of the differential equation $y' = \frac{y^2 + 1}{xy + y}$? I start out with the separable differential equation, 
$$y' =\frac{dy}{dx} = \frac{y^2 + 1}{xy + y} = \frac{y^2 + 1}{y(x+1)}$$
Thus, $\frac{1 }{x+1}dx = \frac{y }{y^2 + 1}dy$.
Then integrating both sides of the equation, I get 
$$\ln(x+1) = \frac{1}{2}\ln(y^2 +1) + C$$
Now, $e^{\ln(x+1)}$ = $e^{\frac{1}{2}\ln(y^2 +1) + C}$. So...
$$(x+1) = e^C(y^2 + 1)^{\frac{1}{2}}$$
I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.
 A: You should get $x+1=C_1\sqrt{y^2+1}$, where $C_1=e^C$.
A: You correctly obtained the general solution :
$$x+1=C_1\sqrt{y^2+1} \tag 1$$
Writing it on explicit form for $y(x)$ :
$$y(x)=\pm\sqrt{C_2(x+1)^2-1} \tag 2$$
with $C_2=\frac{1}{C_1^2}$
The result presented on the form $(2)$ forgets the particular case $C_1=0$ which corresponds to the singular solution $x=-1$, any $y$, that is $x(y)=-1$ , which is represented by a vertical straight line in Cartesian coordinates.
The line $x(y)=-1$ is the envelope of the curves of Eq.$(2)$.
A: I believe that $x + 1 = \pm e^C\sqrt{y^2 + 1}$ is the correct general solution. You can rewrite it as: $x + 1 = C\sqrt{y^2 + 1}$, where $C = \pm e^c \neq 0$.
As for your concern about the singular solution, I don't think the original equation has $x = -1$ as the singular solution because $x = -1$ makes the original equation undefined.
If only the original equation is initially given in the form:
\begin{align}y'(x + xy) = y^2 + 1\end{align}
then the solution $x = -1$ does not make the original equation undefined and therefore is singular.
Apologize for bad Engish.
