17. dP/dt=P-P^2 by separable method.

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17. dP/dt=P-P^2 by separable method.

To solve the differential equation \frac{dP}{dt}=P-P^2 by separable method, we need to separate the variables P and t on opposite sides of the equation and integrate both sides.

Starting with the given differential equation:

    \begin{equation*} \frac{dP}{dt} = P-P^2 \end{equation*}

We can divide both sides by (P-P^2) to obtain:

    \begin{equation*} \frac{1}{P-P^2}dP = dt \end{equation*}

Now we can integrate both sides with respect to their respective variables:

    \begin{align*} \int \frac{1}{P-P^2}dP &= \int dt \end{align*}

    \begin{align*} \int \frac{1}{P(1-P)}dP &= t + C_1 \end{align*}

    \begin{align*}\int \left(\frac{1}{P}+\frac{1}{1-P}\right)dP &= t + C_1 \end{align*}

    \begin{align*}\ln|P| - \ln|1-P| &= t + C_2 \end{align*}

    \begin{align*}\ln\left|\frac{P}{1-P}\right| &= t + C_2 \end{align*}

    \begin{align*}\frac{P}{1-P} &= Ce^{t} \end{align*}

where C is a constant determined by the values of the constants of integration C_1 and C_2.

Simplifying the last equation, we get:

    \begin{align*} P &= \frac{Ce^t}{1+Ce^t} \ \end{align*}

where C is a constant determined by the initial conditions.

Therefore, the general solution to the given differential equation is:

    \begin{equation*} P = \frac{Ce^t}{1+Ce^t} \end{equation*}

where C is a constant determined by the initial conditions.

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