Question

Find the explicit solution for: dX/dt=(x-1)(2x-1), ln(2x-1/x-1)=t

Answer 1

$\dfrac{\mathrm dx}{\mathrm dt}=(x-1)(2x-1)$

is a separable ODE, as

$\dfrac{\mathrm dx}{(x-1)(2x-1)}=\mathrm dt$

Decompose the left side into partial fractions:

$\dfrac1{(x-1)(2x-1)}=\dfrac1{x-1}-\dfrac2{2x-1}$

Then integrating both sides gives

$\displaystyle\int\left(\frac1{x-1}-\frac2{2x-1}\right)\,\mathrm dt=\int\mathrm dt$

$\ln|x-1|-\ln|2x-1|=t+C$

Solve for $x(t)$:

$\ln\left|\dfrac{x-1}{2x-1}\right|=t+C$

$\dfrac{x-1}{2x-1}=e^{t+C}=Ce^t$

$x-1=(2x-1)Ce^t$

$x(1-2Ce^t)=1-Ce^t$

$\implies\boxed{x(t)=\dfrac{1-Ce^t}{1-2Ce^t}}$