$(|x|=|y|\implies x=y)$ while solving $\frac{\mathrm dy}{\mathrm dx}=\frac{y}{1+x}$

Question

G Zill, Dennis, Differential Equations with Boundary-Value Problems, 7th edition, p. 46:

Example 1:

Solve $(1+x)\mathrm dy-y\mathrm dx=0.$

$$\frac{\mathrm dy}{\mathrm dx}=\frac{y}{1+x}\\\cdots\\\cdots$$

From $$\ln|y|=\ln|c(1+x)| \tag{1}$$ we immediately get

$$y=c(1+x)\tag{2}$$

Given that $x,y\in\mathbb{R}$ and $|x|=|y|,$ I don't think we can say that $x=y.$ Why did the author go from $(1)$ to $(2)$ ? Is that step mathematically rigorous?

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EDIT

Follow-up question: How to solve for $y$ in $(1+x)dy-ydx=0$?

Answer 1

$y=\pm c(1+x)$. But, as solution of the DE, $y$ is a differentiable function. This forces $y$ to be $ c(1+x)$ for all $x$ or $- c(1+x)$ for all $x$. In the second case, the minus sign can be absorbed inside the constant $c$.

Answer 2

1. $$\ln|y|=\ln|c(1+x)| \tag{1}$$ $$y=c(1+x)\tag{2}$$

   Why did the author go from (1) to (2) ?

   \begin{align} &\ln|y|=\ln|C_1(1+x)| \\ \iff{}&|y|=|C_1(1+x)|\quad\land\quad y\ne0\\ \iff{}&y=\pm C_1(1+x) \quad\land\quad y\ne0\\ \iff{}&y=C_2(1+x) \quad\land\quad y\ne0 \end{align}

   Notice that $C_1$ and $C_2$—like the expression $\pm e^{C}$—are arbitrary nonzero constants.

2. Suppose that $y=f(x)$ on $\mathbb R.$ Then, since the starting differential equation is $$\boxed{\frac{\mathrm dy}{\mathrm dx}=\frac{y}{1+x}},\tag{*}$$ the arbitrary constants $A, B$ and $D$ in its general solution (where $\boldsymbol{A\ne B}$ and $\boldsymbol{E\ne 0}$) $$\boxed{y=\begin{cases} A(1+x), &x<-1 \\ D, &x=-1 \\ B(1+x), &x>-1\end{cases}\quad\text{or}\quad y=\begin{cases} D(1+x), &x<-1 \\ E, &x=-1 \\ D(1+x), &x>-1\end{cases}}\tag{*}$$ can be zero. For example, $$y=\begin{cases} -7(1+x), &x<-1 \\ 0, &x=-1 \\ 3(1+x), &x>-1\end{cases}$$ and $$y=\begin{cases} 0, &x<-1 \\ -2, &x=-1 \\ 0, &x>-1\end{cases}$$ are two particular solutions.

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Reply to comment

How is $|y|=\pm y$? Isn't $|y|=\begin{cases} y, y>0\\-y, y<0 \end{cases}$?

$|y|\equiv±y$ is a false identity because $|y|$ and $\pm y$ are not generally subtitutable for each other, and we did not invoke it. What we invoked above is this equivalence: $$|m|=|n|⟺m=±n.$$ Notice that $|3|=|−3|$ and $|3|=|3|$ and $|−3|=|−3|.$

This may be of interest: How do you read $±$? Why does $|x|=3⟹x=±3$?