# Integral of two-variable function

1. May 15, 2012

### Allen3

I'm working on a problem where the current i is described by the equation:

di/dt = i*A/(B+C*t)

The rate of change in i therefore depends on both i and t. I'm looking for a closed-form solution for i(t) given i(0)=I. I have to admit my calculus is pretty rusty and I'm having trouble figuring out how to solve that equation.

Can anyone offer some assistance or a push in the right direction? Thanks much.

2. May 15, 2012

### Allen3

Forget it--I figured it out

OK, my apologies. I was able to solve the problem using variable substitution--basically, finding di/dC and then finding i(C). Sorry for the false alarm.

3. May 15, 2012

### arildno

C is a constant???

4. May 15, 2012

### HallsofIvy

??? You say that the rate of change depends upon both i and t. So where does "C" come into this? Given that
$$\frac{di}{dt}= \frac{Ai}{B+ Ct}$$
a "separable equation", you can rewrite it as
$$\frac{di}{i}= \frac{Adt}{B+ Ct}$$
which can be integrated.

5. May 16, 2012

### Allen3

Yes, C is a constant. Looking back at my notes, it looks like what I did was to set q(t) = B+C*t, then di/dt = i*A/q, and since dq/dt = C, (di/dt)*(dt/dq) = di/dq = i*A/(C*q) and I then integrated (1/i)*di = (A/C)*(1/q)*dq to find i(q) and made some substitutions to find i(t).

But back to the original problem, where you left it is where I was stumped: what is the integral of A*dt/(B+C*t) ?

6. May 16, 2012

### HallsofIvy

That's a fairly standard "Calculus II" problem. Let u= B+Ct so that du= Cdt and dt=du/C.
$$\int \frac{Adt}{B+ Ct}= A\int \frac{du/C}{u}= \frac{A}{C}\int \frac{du}{u}$$

7. May 16, 2012

### Allen3

Ah, thanks. For some reason I have no specific memory of learning that, but apparently the lesson soaked in somehow and helped me figure it out. Thanks.