# Why does the absolute value disappear when taking $e^{\ln|x|}$

I have noticed that if you have an equation (after integrating) such as $$\ln|y| = \ln|x| + c,$$ and you further simplify it using the law of exponents, you get $$e^{\ln|y|} = e^{\ln|x|+с},$$ which is the same as $y = cx$.

My question is why the absolute value disappears all of a sudden.

Edit: The original questions is:

Solve the separable differential equation: $(1+x)dy - ydx = 0$.

Final solution: $y = c(1+x)$.

• I have updated the post to LaTeX, please see that it is correct. – Jeel Shah Oct 2 '13 at 0:13
• Why do you think it does? Seems like it should be $|y|=K|x|$ where $K=e^c$. Who says otherwise? – Thomas Andrews Oct 2 '13 at 0:18
• You are correct it should have been Kx I just updated my post, but I've seen the absolute value being ignored in the book that I'm currently reading. – fYre Oct 2 '13 at 0:22
• The thing is that $|y|=k|x|$ is not the graph of a function. You just show that your solution's graph is contained in $|y|=k|x|$. Then you see what subset of it is a solution. – OR. Oct 2 '13 at 0:22
• Please note I do know that it is a different constant everytime I make a a change to the original constant, I just used c over and over again for simplicity, sorry for any confusion. – fYre Oct 2 '13 at 0:34

You're getting confused because you're using $c$ for three different purposes, but thinking they're all the same.

The equation $$e^{\ln |y|} = e^{\ln |x| + c}$$ does not simplify to $cx$ or even $c|x|$ on the right hand side: it simplifies to $$|y| = e^c |x|$$ We could, however, introduce a new constant $c_2 := e^c$, so that we can write $$|y| = c_2 |x|$$

Now, we can use the laws of absolute values to break apart this equation:

$$\begin{cases} y = c_2 x & x \geq 0, y \geq 0 \\ y = -c_2 x & x \leq 0, y \geq 0 \\ -y = c_2 x & x \geq 0, y \leq 0 \\ -y = -c_2 x & x \leq 0, y \leq 0 \end{cases}$$

This simplifies to

$$\begin{cases} y = c_2 x & x \geq 0, y \geq 0 \text{ or } x \leq 0, y \leq 0 \\ y = -c_2 x & x \leq 0, y \geq 0 \text{ or } x \geq 0, y \leq 0\end{cases}$$

If we want $y$ to be expressed as a continuous function of $x$, there are four ways to define $y = f(x)$ so that this is true:

$$f(x) = c_2 x$$ $$f(x) = -c_2 x$$ $$f(x) = c_2 |x|$$ $$f(x) = -c_2 |x|$$

If you require $f$ to be differentiable, then only the first two are possible.

So you set $c_3 := c_2$ or $c_3 := -c_2$ as appropriate. Then the solution reduces to

$$y = c_3 x$$

(note that because the set of possible values for $c_1$ ranges over all real numbers, the set of possible values for $c_3$ ranges over all nonzero real numbers)

An example where this occurs is solving the differential equation $\dfrac{dy}{dx} = \dfrac{y}{x}$ which goes like this

\begin{align*} \frac{1}{y}\frac{dy}{dx} &= \frac{1}{x}\\ \int\frac{1}{y}\frac{dy}{dx}dx &= \int\frac{1}{x}dx\\ \int\frac{1}{y}dy &= \ln|x|+c\\ \ln|y| &= \ln|x|+c\\ e^{\ln|y|} &= e^{\ln|x|+c}\\ |y| &= K|x| \end{align*}

where $K = e^c > 0$. Note, for a fixed $x_0$ in the domain (which is a subset of $\mathbb{R}\setminus\{0\}$), we have $y = K|x_0|$ or $y = -K|x_0|$. As we are solving a differential equation, $y$ is differentiable and hence continuous. Therefore, if $y = -K|x_0|$ for some fixed $x_0$, we must have $y = -K|x|$ for all $x$ in a neighbourhood of $x_0$ (likewise if $y = K|x_0|$).

Suppose now that the domain is connected - $\mathbb{R}\setminus\{0\}$ is not connected but $(-\infty, 0)$ and $(0, \infty)$ are, they are the connected components of $\mathbb{R}\setminus\{0\}$ (the largest connected subsets of $\mathbb{R}\setminus\{0\}$). Then we must have $y = K|x|$ or $y = -K|x|$ on the entire domain. Note, we can combine these two families of solutions into one: $y = A|x|$ where $A \in \mathbb{R}\setminus\{0\}$. However, the domain is connected if and only if it is a subset of $(-\infty, 0)$ or $(0, \infty)$. Depending on which of the two sets the domain is contained in, we either have $|x| = -x$ or $|x| = x$, so that the family of solutions can be written as $y = -Ax$ for $A \in \mathbb{R}\setminus\{0\}$ or $y = Ax$ for $A \in \mathbb{R}\setminus\{0\}$. Again, we can combine these two families of solutions into one: $y = Bx$ where $B \in \mathbb{R}\setminus\{0\}$. Note, this family gives all the solutions on any connected domain. If the domain is not connected, we have to consider potentially different solutions on each connected component. For example,

$$y = \begin{cases} x &\ \text{if}\ x > 0\\ -x &\ \text{if}\ x<0 \end{cases}$$

is a solution to the differential equation but is not of the form $y = Bx$ for some universal constant $B$; it is however of the form $y = B(x)x$ for some locally constant function $B$.

This is wrong because $e^{\ln|x| + c} = e^c|x|$. But if your question states that $x \geq 0$, then $e^{\ln|x| + c} = e^cx = Cx$, where $C = e^c$. Also note that $e^\xi> 0$ for all real $\xi$, so $e^{\ln|x| + c} > 0$.

• The question says solve the separable differential equation: (1+x)dy - ydx = 0. Maybe I should just assume x >= 0? – fYre Oct 2 '13 at 0:27
• Also, the final solution is: y = c(1+x). – fYre Oct 2 '13 at 0:30