Removal of absolute signs I'm a high school student and I'm currently studying differential equations. I encountered this one question:
An object is dropped from a cliff. The object leaves with zero speed, and t seconds later its speed v metres per second satisfies the differential equation
$$
\frac{dv}{dt}=10-0.1v^2
$$
So I found t in terms of v
$$
t=\frac{1}{2}\ln\left|\frac{10+v}{10-v}\right|
$$
The questions goes on like this: Find the speed of the object after 1 second. Part of the answer key shows this
$$
t=\frac{1}{2}\ln\left|\frac{10+v}{10-v}\right|
$$
$$
2t=\ln\left|\frac{10+v}{10-v}\right|
$$
$$
e^{2t}=\frac{10+v}{10-v}
$$
So here's my question: why is it not like this?
$$
\pm e^{2t}=\frac{10+v}{10-v}
$$
Why can you ignore the absolute sign? Thanks!
 A: You don't ignore the sign. Let's look at the original equation, and notice what's happening at $v=10$. The acceleration is zero: $$\frac{dv}{dt}=10-0.1\cdot 10^2=0$$
Since you started with $v=0$, you can see that the object accelerates, initially with $dv/dt=10$, but then the acceleration starts to decrease, as the object approaches $v=10$. After that the velocity can't increase any more. So with your initial condition, you have $v<10$. Then the sign of the expression in the absolute value is always positive. Note that if you would have started with an initial velocity greater than $10$, then the acceleration would have been negative, but always $v>10$ (getting close to $10$ as $t\to\infty$). Then you would use the negative sign when you explicitly write the absolute value.
A: In general what you wrote is correct,
$$
t+c=\ln\left|\frac{10-v}{10+v}\right|
$$
has to be transformed to
$$
\frac{10-v}{10+v}=\pm e^ce^t=Ce^t,
$$
where $C\ne 0$ in this context. However $C=0$ and $C=\infty$ correspond to the previously excluded (for the separation approach) constant solutions.
$C$ is a constant over all the solution, its value and thus its sign gets determined by the initial condition. At $t_0=0$ this gives
$$
C=\frac{10-v_0}{10+v_0}
$$
