Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integral of two-variable function

  1. May 15, 2012 #1
    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. jcsd
  3. May 15, 2012 #2
    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.
     
  4. May 15, 2012 #3

    arildno

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    C is a constant???
     
  5. May 15, 2012 #4

    HallsofIvy

    User Avatar
    Science Advisor

    ??? You say that the rate of change depends upon both i and t. So where does "C" come into this? Given that
    [tex]\frac{di}{dt}= \frac{Ai}{B+ Ct}[/tex]
    a "separable equation", you can rewrite it as
    [tex]\frac{di}{i}= \frac{Adt}{B+ Ct}[/tex]
    which can be integrated.
     
  6. May 16, 2012 #5
    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) ?
     
  7. May 16, 2012 #6

    HallsofIvy

    User Avatar
    Science Advisor

    That's a fairly standard "Calculus II" problem. Let u= B+Ct so that du= Cdt and dt=du/C.
    [tex]\int \frac{Adt}{B+ Ct}= A\int \frac{du/C}{u}= \frac{A}{C}\int \frac{du}{u}[/tex]
     
  8. May 16, 2012 #7
    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook