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Annette [7]
3 years ago
14

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

Mathematics
1 answer:
LiRa [457]3 years ago
8 0

\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}}

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