# Math Help - Understanding notation in a 1st order linear differentiation equation

1. ## Understanding notation in a 1st order linear differentiation equation

Hi all,

I'm having difficulty understanding conventions for DE, and I suppose calculus in general. I have not had calculus in several years, so I'm trying to figure out what exactly is happening in this sequence.

Here's what the book says:
$(du(t)/dt) = (1/2) (u(t))$
$(du(t)/dt) / u(t) = 1/2$
$(d/dt) ln|u(t)| = 1/2$
$ln |u(t)| = (1/2)t + C$
$u(t) = ce^(t/2)$

So, I understand the basics: rearrange the equation, integrate both sides, and solve for u(t).

My only concern is that I have no idea what the derivative notation means. For example, what exactly is happening from line 2 to line 3 that makes the du(t) drop out. Does this mean that when I integrate (du(t)/dt) (1/ u(t)) that du drops out? Ok, but why is the (d/dt) left? Doesn't that mean "take the derivative"?

$(d/dt) INTEGRAL((1/ u(t)) d(u) = 1/2$

^Is that what's happening? If so, then I still don't understand the next steps. Where is that d/dt going exactly??

I hope you can understand what I'm asking: what do all these d's, dt's, and d(u)'s mean. I usually just ignore them and remember the process.

Thanks.

2. This is sloppy working as it skips steps.

I'd do it like this...

$\frac{du}{dt} = \frac{1}{2}u$

$\frac{1}{u}\,\frac{du}{dt} = \frac{1}{2}$

$\int{\frac{1}{u}\,\frac{du}{dt}\,dt} = \int{\frac{1}{2}\,dt}$

$\int{\frac{1}{u}\,du} = \int{\frac{1}{2}\,dt}$

$\ln{|u|} + C_1 = \frac{1}{2}t + C_2$

$\ln{|u|} = \frac{1}{2}t + C_2 - C_1$

$|u| = e^{\frac{1}{2}t + C_2 - C_1}$

$|u| = e^{C_2 - C_1}e^{\frac{1}{2}t}$

$u = \pm e^{C_2 - C_1}e^{\frac{1}{2}t}$

$u = Ce^{\frac{1}{2}t}$ where $C = \pm e^{C_2 - C_1}$.

3. Thanks for explaining in detail!

On the third line, are the dt's crossing out? Is that what's actually happening?

4. It's an application of the chain rule in reverse. It's almost like the $dt$s "cancel", but technically speaking you can't say that they cancel since a derivative is not a fraction, it only works like a fraction.