Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given $K(0) = 0,2P$. I'm supposed to solve the ODE

$$ \frac{dK}{dt} = \lambda K(P-K)$$

I have tried to seperate and integrate both sides

$$ \int \frac{1}{K(P-K)} dK = \int \lambda \space dt$$

to get

$$ \ln|K(P-K)| = \lambda t + C$$

and then solve for $K$

$$ e^{\ln|K(P-K)|} = K(P-K)=e^{\lambda t + C}$$

But there I'm stuck as to getting any further to finding the general solution. Does the $K(P-K)$ term require integrating using partial fractions?

share|cite|improve this question
Yes, you need partial fractions. What you wrote after your integral is incorrect. In particular, $$ \frac{1}{P} \left[ \frac{1}{K} + \frac{1}{P-K} \right] = \frac{1}{K(P-K)} $$. – Christopher A. Wong Oct 31 '12 at 0:04
At the point at which you say you are stuck, you have a quadratic equation for $K$. Surely you can solve a quadratic equation? You'll still need to do this once you've corrected the integration. – Gerry Myerson Oct 31 '12 at 0:14
up vote 1 down vote accepted

Yes, you need partial fractions. See this question about partial fractions. Write $$\frac{1}{{K(P - K)}} = \frac{a}{K} + \frac{b}{{(P - K)}}$$ and solve for $a$ and $b$. If the link does not help, Google it. It is not as difficult as it might seem, especially in this case.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.