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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 $C$

Thanks

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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

2 Answers 2

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.

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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 $X(0)$:

$$X(0)=e^Ce^0=e^C$$

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 $C$.

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