How to solve $\frac{\mathrm{d}I}{\mathrm{d}t} = (\beta - \gamma)I$?

Question

I need help with solving this exponential growth equation: $$\frac{\mathrm{d}I}{\mathrm{d}t} = (\beta - \gamma)I.$$

Answer 1

If the term $(\beta - \gamma)$ is independent of $t$, then it's the same case as solving the differential equation $\dfrac{\mathrm{d}I}{\mathrm{d}t} = aI$ where $a \in \mathbb R$ is a constant. Thus you just separate (in the case of $I \neq 0)$:

$$\dfrac{\mathrm{d}I}{\mathrm{d}t} = (\beta - \gamma)I \Leftrightarrow \frac{1}{I}\mathrm{d}I = (\beta-\gamma)dt \Rightarrow \int \frac{1}{I}\mathrm{d}I = \int(\beta-\gamma)dt $$ $$\implies$$ $$\ln I(t) = (\beta - \gamma)t + C \Leftrightarrow I(t) = C'e^{(\beta-\gamma)t}$$ where $C \in \mathbb R$ is the indefinite integration consant and $C' = e^C$. This can be calculated by some initial conditions.

Answer 2

Assuming $\beta$ and $\gamma$ are constant with respect to t, then:

$\frac{I'}{I}=\beta -\gamma \to ln(I)=(\beta-\gamma)t+C \to I=ce^{(\beta-\gamma)t}$

Answer 3

If you have problems to solve this kind of first order equations (separable equation), tthis easy-rule will help you: you can see the differential operator as a "fraction" $$\dfrac{dI}{dt} = (\beta-\gamma)I$$

$$\frac{1}{I}dI = (\beta-\gamma)dt$$

$$\displaystyle\int\frac{1}{I}dI = \int(\beta-\gamma)dt$$ $$\ln(I) = (\beta-\gamma)t+C$$ $$I = K\cdot e^{(\beta-\gamma)t}$$

Answer 4

If $β$ and $\gamma$ depend on time, with the same method as Rebellos, you would find $$ \ln I(t) = \ln I(0) + \int_0^t \beta(s)-\gamma(s)\,\mathrm{d}s, $$ so that at the end you get $$ I(t) = I(0)\, e^{\int_0^t \beta(s)-\gamma(s)\,\mathrm{d}s}. $$