How do I find the singular solution of the differential equation $y' = \frac{y^2 + 1}{xy + y}$?

Question

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.

Answer 1

You should get $x+1=C_1\sqrt{y^2+1}$, where $C_1=e^C$.

Answer 2

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.$(2)$.

Answer 3

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.