Solve the differential equation it's a modified logistic model

Question

$$\frac{dx}{x\left(1-\frac{x}{a}\right)\left(\frac{x}{b}-1\right)}=cdt$$

i tried to solve this equation by variable separable, but i am not able to do the partial fraction properly

Answer 1

The partial fractions can be written as $$ \frac{1}{x\left(1-\frac{x}{a}\right)\left(\frac{x}{b}-1\right)} = -\frac{1}{x}+{\frac {a}{ \left( a-b \right) \left( x-b \right) }}-{ \frac {b}{ \left( a-b \right) \left( x-a \right) }} $$ so the general solution can be written as $$-\ln|x| + \frac{a}{a-b} \ln|x-b| - \frac{b}{a-b} \ln |x-a| = c t + \text{const} $$

Answer 2

Note that

$$\frac{1}{x(1-x/a)(x/b-1)}=\frac{A}{x}+\frac{B}{x-a}+\frac{C}{x-b} \tag 1$$

Now, multiplying $(1)$ by $x-b$ and taking a limit as $x\to b$ reveals

$$\begin{align} \lim_{x\to b}\left(\frac{x-b}{x(1-x/a)(x/b-1)}\right)&=\lim_{x\to b}\left(\frac{(x-b)A}{x}+\frac{(x-b)B}{x-1}+\frac{(x-b)C}{x-b}\right)\tag 2 \end{align}$$

and we find that $C=\frac{a}{a-b}$.

Can you find $A$ and $B$ similarly now?