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

I never get why the constant of the integration in a ODE is related to a initial condition. for example suppose the following ODE with initial condition $X(0)=x_0$.

$X'(t)=cX(t):\frac{dX}{X}=cdt\Longrightarrow\ln X-\ln x_0=ct\Longrightarrow X(t)=x_0e^{ct}$

Why emerges a $\ln x_0$ and no other ordinary constant $C$


share|cite|improve this question
It is an ordinary constant $C$. Just happened to be that it's also equal to the initial condition. Can be validated by simple substitution. – Kaster Jun 12 '13 at 21:37

You can use an ordinary constant $C$:

$$\ln X=ct+C$$ $$X=e^{C}e^{ct}$$

Now, what's the initial condition? It's the value $X(0)$:


So the initial value in this equation is $e^C$, and so for a matter of notation, we introduce $x_0=e^C$ as the initial value instead of leaving it as a function of $C$.

share|cite|improve this answer

Integrate. We get $$\ln X=ct+C.\tag{1}$$ Before going further, let's evaluate $C$. When $t=0$, we have $X=x_0$. Substituting in Equation (1), we get $$\ln x_0=C,$$ or equivalently $$\ln X-\ln x_0=ct.$$ Now take the exponential of both sides.

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.