91影视

Let the given function be \(P = \frac{{{c_1}{e^t}}}{{1 + {c_1}{e^t}}}\).

The first derivative of the function is,

\(\frac{{dP}}{{dt}} = \frac{{{c_1}{e^t}(1 + {c_1}{e^t}) - {c_1}{e^t}({c_1}{e^t})}}{{{{(1 + {c_1}{e^t})}^2}}}\)

Substitute \(y\) and \(y'\) in the differential equation.

\(\begin{aligned}{c}\frac{{{c_1}{e^t}(1 + {c_1}{e^t}) - {c_1}{e^t}({c_1}{e^t})}}{{{{(1 + {c_1}{e^t})}^2}}} &= \frac{{{c_1}{e^t}}}{{1 + {c_1}{e^t}}}\left( {1 - \frac{{{c_1}{e^t}}}{{1 + {c_1}{e^t}}}} \right)\\\frac{{{c_1}{e^t}}}{{{{(1 + {c_1}{e^t})}^2}}} &= \frac{{{c_1}{e^t}}}{{1 + {c_1}{e^t}}}\left( {\frac{1}{{1 + {c_1}{e^t}}}} \right)\\\frac{{{c_1}{e^t}}}{{{{(1 + {c_1}{e^t})}^2}}} &= \frac{{{c_1}{e^t}}}{{{{(1 + {c_1}{e^t})}^2}}}\end{aligned}\)

Hence, the left-hand limit is equal to the right-hand limit, so the solution is verified.

Case 1: If \({c_1} \ge 0\), then the denominator will always be positive for any real number \(t\). The solution is defined at every real value of \(t\).

Hence, the interval at \({c_1} \ge 0\) is \(I:( - \infty ,\infty )\).

Case 2: If \({c_1} < 0\), there is a chance that the denominator will be zero. So,

\(\begin{array}{c}1 + {c_1}{e^t} \ne 0\\{c_1}{e^t} \ne - 1\\{e^t} \ne \frac{{ - 1}}{{{c_1}}}\\t \ne ln\left( {\frac{{ - 1}}{{{c_1}}}} \right)\end{array}\)

Hence, the interval at \({c_1} < 0\) is \(I:\left( { - \infty ,ln\left( {\frac{{ - 1}}{{{c_1}}}} \right)} \right)or\left( {ln\left( {\frac{{ - 1}}{{{c_1}}}} \right),\infty } \right)\).