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

Say I have a value which is growing at a constant percentage-increase. I can calculate this using the exponential growth function, where $r$ is my growth rate; $y=y_0(1+r)^t$ Now, consider the case where $r$ is not constant, but rather changes with time: $r(t)$. For example, $r(t)$ could be the return on investment for an asset which varies with market conditions. If I know the integral of $r(t)$, can I use this to calculate $y$? How would I go about this?

share|cite|improve this question
up vote 0 down vote accepted

Suppose we have discrete time: Then $y(t) = y_0 \prod_{i=0}^{t-1}(1+r(i)$).

The continuous analogue is, as far as I recall: $y(t) = y_0 \exp(\int_0^t r(x) dx)$.

(Somebody correct me if I'm wrong; at least it makes sense for $r(i) = r$ constant.)

share|cite|improve this answer

If $f(t)$ grows exponentially, that is if the growth (or derivative) of $f$ at some point $t$ is proportional to $f(t)$, $f$ is described by the following linear differential equation with constant coefficients $$ f'(t) = \lambda f(t) $$

If you replace the constant $\lambda$ with a function $\lambda(t)$ you again get a linear differential equation, but with time-varying coefficients $$ f'(t) = \lambda(t) f(t) $$

Equations of that type can always be solved by separation of variables ( ). If you set $y = f(t)$, you get $$ \frac{dy}{dt} = \lambda(t) y $$ Then you throw all caution to the wind, forget that you're not supposed to handle $dx$ and $dy$ as if they were actual quantities, and get $$ \begin{eqnarray*} \frac{1}{y} dy &=& \lambda(t) dt &\implies\\ \int \frac{1}{y} dy &=& \int \lambda(t) dt &\implies\\ f(t) &=& e^{\int \lambda(t) dt} \end{eqnarray*} $$

By the chain rule, $f'(t) = \lambda(t) e^{\int_0^t \lambda(t) dt}$, i.e. $f$ is actually a solution of the original differential equation. Since you can pick any antiderivative of $\lambda(t)$, i.e. add an arbitrary constant to $\int \lambda(t) dt$, you might was well pick $\int_0^t \lambda(t) dt$, which yields the general solution $$ f(t) = f(0)e^{\int_0^t \lambda(t) dt} $$

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.