Question

Verify that P = Ce^t /1 + Ce^t is a one-parameter family of solutions to the differential equation dP dt = P(1 − P).

Answer 1

Answer:

See verification below

Step-by-step explanation:

We can differentiate P(t) respect to t with usual rules (quotient, exponential, and sum) and rearrange the result. First, note that

$1-P=1-\frac{ce^t}{1+ce^t}=\frac{1+ce^t-ce^t}{1+ce^t}=\frac{1}{1+ce^t}$

Now, differentiate to obtain

$\frac{dP}{dt}=(\frac{ce^t}{1+ce^t})'=\frac{(ce^t)'(1+ce^t)-(ce^t)(1+ce^t)'}{(1+ce^t)^2}$

$=\frac{(ce^t)(1+ce^t)-(ce^t)(ce^t)}{(1+ce^t)^2}=\frac{ce^t+ce^{2t}-ce^{2t}}{(1+ce^t)^2}=\frac{ce^t}{(1+ce^t)^2}$

To obtain the required form, extract a factor in both the numerator and denominator:

$\frac{dP}{dt}=\frac{ce^t}{1+ce^t}\frac{1}{1+ce^t}=P(1-P)$