7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ I have updated the post to LaTeX, please see that it is correct. $\endgroup$ – Jeel Shah Oct 2 '13 at 0:13
  • $\begingroup$ Why do you think it does? Seems like it should be $|y|=K|x|$ where $K=e^c$. Who says otherwise? $\endgroup$ – Thomas Andrews Oct 2 '13 at 0:18
  • $\begingroup$ 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. $\endgroup$ – fYre Oct 2 '13 at 0:22
  • $\begingroup$ 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. $\endgroup$ – OR. Oct 2 '13 at 0:22
  • 3
    $\begingroup$ 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. $\endgroup$ – fYre Oct 2 '13 at 0:34
8
$\begingroup$

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)

$\endgroup$
3
$\begingroup$

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 $B$.

$\endgroup$
0
$\begingroup$

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 $e^{\ln|x| + c} > 0$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.