Q:

How do you solve the following separable differential equation: y'y = y + 1?

I just started learning about differential equations and encountered following equation:
$$ y'y = y +1 $$
Wolfram alpha provided the following explanation:
here
But I'm not sure how the integration is performed. What rules are used?
How does $\int{\frac{y'y}{y+1}dx}$ become $-\log(y + 1) + y$?

A:

They use substitution, namely that $y'dx = dy$, so you can rewrite
$$
\int\frac{y'y}{y+1}dx = \int\frac{y}{y+1} dy = \int\left(\frac{-1}{y+1} + 1\right)dy
$$
Using this substitution to solve a differential equation is called "solution by separation", and an equation where this is possible to do is called a "separable equation". The general requirement for a separable equation is that you can get the equation to the form $f(y)y' = g(x)$, which then becomes $\int f(y) dy = \int g(x)dx$.

A:

Note that Wolfram Alpha is wrong. The correct real-valued solutions are $x = y - \ln|y+1| + c$ for some constant $c$, or $y = -1$. The reason it is missed the latter is that the technique of separable variables (equivalent the substitution rule) is applied wrongly. One can only carry out that solution where $y \ne -1$. Then one needs to check that those regions of the solution do not touch $y = -1$ and hence there is no bifurcation. See this post for an example with bifurcation. In general Wolfram Alpha always gets this kind of differential equation wrong, though arguably in the real world functions are sufficiently smooth and so the strange solutions are never physical.
Here's the proper solution:
$\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

Take any real variables $x,y$ such that $\lfrac{dy}{dx} y = y+1$.
Let $D$ be any open subset of the curve defined by $(x,y)$ that does not intersect the line defined by $y+1=0$.
Then $\lfrac{dy}{dx} \lfrac{y}{y+1} = 1$ on $D$.
Thus $\int \lfrac{dy}{dx} \lfrac{y}{y+1}\ dx = \int 1\ dx = x+k$ on $D$, for some constant $k$.
Also $\int \lfrac{y}{y+1} \lfrac{dy}{dx}\ dx = \int \lfrac{y}{y+1}\ dy$ on $D$, because $\lfrac{y}{y+1}$ is continuous on $D$.
And $\int \lfrac{y}{y+1}\ dy = \int (1-\lfrac{1}{y+1})\ dy = y-\ln|y+1| + m$ on $D$, for some constant $m$.
Thus $x = y-\ln|y+1| + c$ on $D$, for some constant $c$.
Note that $x = y-\ln|y+1|+c \to \infty$ as $y \to -1$, and hence the above solution on $D$ cannot be extended to any solution that includes a point on the line defined by $y=-1$.
Thus there is no bifurcation.
Also note that if $y = -1$ at any point then $\lfrac{dy}{dx} = 0$ and hence $y$ is constant there.
Thus the only other solution is $y = -1$.

A:

$$\frac{dy}{dx}y=y+1$$
$$\int\frac{y}{y+1}dy=\int dx$$
$$\int1-\frac{1}{y+1}dy=\int dx$$
$$y-\log(y+1)=x+c$$

