# Looking for literature on a basic topic

1. Aug 21, 2013

### mindheavy

I've taken differential and integral calculus, and have a fair understanding of the material at this point. Last semester I had a professor show something in class that I feel is a pretty basic idea, I just haven't thought about working things in such a way.

He put a derivative on the board such as the one for instantaneous current in a circuit. He went on to say "you may have not seen this before or knew that you are allowed to do this" and rearranged the formula to then integrate both sides and end up with a new equation.

It is this method I'm interested in. I wonder if there is any material out there that explains this way of rearranging a derivative for example and coming up with new equations by integrating both sides, it's a newish idea to me and want to read more about how or why that works, the motive of it.

I haven't had much luck searching for any literature on it, maybe I don't know how to phrase the search term well enough.

2. Aug 21, 2013

### verty

It's called separation of variables and is taught in any ODE class, and sometimes toward the end of calculus books. I don't know where you could read about the motivation though, but how to do it, for sure.

3. Aug 21, 2013

### mindheavy

Thanks, this explains why I don't know much about it; I'm just now starting vector calculus, am taking Diff. Eq. next semester.

I have a name to go by now, separation of variables, thanks!

4. Aug 21, 2013

### Boorglar

This sounds to me like he solved a differential equation.

Differential equations are equations involving the derivatives of an unknown function, often with some initial or boundary conditions, and solving them means to find the unknown function that fits the additional conditions.

For example, I would suspect your teacher did something like that (this is the differential equation for an RC circuit with no input voltage, like a discharging capacitor):
$R\frac{dQ}{dt} + \frac{Q}{C} = 0 \\ R\frac{dQ}{dt} = -\frac{Q}{C} \\$
divide by Q: $\frac{1}{Q}\frac{dQ}{dt} = -\frac{1}{RC}\\$
integrate both sides: $\int_{t_0}^{t}\frac{1}{Q}\frac{dQ}{dt}dt = \int_{t_0}^{t}-\frac{1}{RC}dt$
use chain rule: $\frac{1}{Q}\frac{dQ}{dt} = \frac{d}{dt}(\ln(Q))\\ \int_{t_0}^{t}\frac{d}{dt}(\ln(Q))dt = \int_{t_0}^{t}-\frac{1}{RC}dt = -\frac{t-t_0}{RC}$
use Fundamental Theorem of Calculus:$\int_{t_0}^{t}\frac{d}{dt}(\ln(Q))dt = \ln(Q)-\ln(Q_0) = \ln(\frac{Q}{Q_0})$

so $\ln(\frac{Q}{Q_0}) = -\frac{t-t_0}{RC}$
$Q = Q_0e^{-\frac{t-t_0}{RC}}$

5. Aug 21, 2013

### mindheavy

thanks Boorglar! I will work through this when I have some time later, this is starting to give me a bit of a grasp of differential equations, seems like the diff eq class will be very useful!

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