Solve $(1+x) d y-y d x=0$ Here is the worked example by the author,
$$
\int \frac{d y}{y}=\int \frac{d x}{1+x}
$$
$$
\ln |y|=\ln |1+x|+c_{1}
$$
$$
|y|=e^{\ln |1+x|+c_{1}}=e^{\ln |1+x|} \cdot e^{c_{1}}=|1+x| e^{c_{1}}
$$
$$
y=\pm e^{c_{1}}(1+x)
$$
Relabeling $\pm e^{c_{1}}$ as $c$ then gives $y=c(1+x)$.
My question: Shouldn't you say $c=\pm e^{c_1},0$ because $y=0$ is a constant solution?
I guess that when $x=-1,y=0$ so I thought maybe that covers the solution $y=0$, but $(1+x) d y-y d x=0$ can be rearranged to $\frac{d y}{d x}=\frac{y}{1+x}$ and this implies that $x \neq-1$, so the constant solution $y=0$ cannot be achieved from $y=c(1+x)$ where $c=\pm e^{c_1}$, is what I think.. Thanks!
 A: When dividing by $y$ you are assuming that $y$ is not the zero function – it may be zero at isolated points. So what happens at $x=-1$ does not matter for this problem. It turns out that the $y=0$ case can be subsumed into the general solution, since it fills the gap left by the expression $c=\pm e^{c_1}$ (which implies $c$ may be any real number except zero). Hence $c$ may be any real number.
A: The procedure followed to solve the equation assumes $y\neq 0$ and $x\neq -1$. Assuming that you get solutions
$$
y=\pm e^{c_1}(1+x)
$$
which is the same as $y=C(1+x)$ with $C\neq 0$. But you realize that $y=0$ is also a solution when the equation is written in the form
$$
\frac{dy}{dx}=\frac y{x+1}
$$
and $x=-1$ is a solution when it is written in the form
$$
\frac{dx}{dy}=\frac{x+1}{y}
$$
You can include the first special solution by allowing $C =0$ and the second one by allowing $C=\infty$ (or by shifting the constant to the other side and allowing it to be. zero).
The differential form of an ODE is more symmetrical and admits solutions which are not necessarily functional relations of the form $y=f(x)$ or $x=g(y)$, but are locally made up of those, except for the points where both coefficients in front of $dx$ and $dy$ vanish (no direction field defined).
