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solve for f: $f' = \alpha\beta f^2-\beta f, 0<\alpha<1, 0<f<1 $ with initial condition $f(0)=1$

Thanks for the hint.

The question has been solved. My working:

Separation of variables is used to solve this differential equation.

$\int\frac {df}{f(\alpha f-1 )}=\beta dt$

To evaluate the integral on the left side, we decompose the fraction into partial fractions $\frac{1}{f(\alpha f-1)}=\frac{\alpha}{\alpha f-1}-\frac1f$. Thus we have $\int (\frac{\alpha}{\alpha f-1}-\frac1f)df=\int\beta dt$

$ln|\frac{\alpha f-1}{f}|=\beta t+C$

$\frac {\alpha f-1}{f} = -e^C\cdot e^\beta t$

...

$f=\frac{1}{\alpha-\alpha e^{\beta t}+e^{\beta t}}$

workinghard
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1 Answers1

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To separate the variables, we want everything involving $f$ or $f'$ on one side, and everything directly involving $t$ on the other. But there's nothing directly involving $t$, so this is easy:

$$ \frac{f'}{\alpha f^2 - f} = \beta $$

Integrating this equation can use partial fraction decomposition:

$$ \frac{1}{\alpha f^2 - f} = \frac{1}{f (\alpha f - 1)} = \frac{A}{f} + \frac{B}{\alpha f - 1} = \frac{A(\alpha f-1) + Bf}{\alpha f^2-f} $$

Can you finish from here?

aschepler
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