banner

Solve the differential equation dp/dt=p−p^2

In this question, we have to find the Integration of the given function $ \dfrac{dP}{dt}= \left[P – P^{2} \right] $ by rearranging the equation.

The basic concept behind this question is the knowledge of derivatives, integration, and the rules such as the product and quotient rules of integration.

Expert Answer

Given function:

\[\dfrac{dP}{dt}= \left[P – P^{2} \right] \]

First, we will rearrange the given equation with $P $ on one side of the equation and $t $ on the other. For this, we have the following equation:

\[dP = \left[P – P^{2} \right] {dt} \]

\[\dfrac{1 }{\left[P – P^{2} \right]} dP = dt \]

\[ dt =\dfrac{1 }{\left[P – P^{2} \right]} dP \]

Take Integration on both sides of the equation. We get:

\[ \int dt = \int \dfrac{1 }{P – P^{2}} dP \]

Taking $P $ common on the right-hand side, we will have the equation:

\[ \int dt = \int \dfrac{1 }{P (1 – P)} dP\]

As we can write $ 1 = ( 1-P ) + P $ in the above equation, putting it in the question we have the following equation:

\[ \int dt = \int \dfrac{(1-P) + P }{P (1 – P)} dP \]

\[ \int dt = \int \dfrac{(1-P) }{P (1 – P)} dP + \int \dfrac{P }{P (1 – P)}  dP \]

Canceling $ 1-P$ from the denominator and numerator of the equation:

\[ \int dt = \int \dfrac{1 }{P } dP + \int \dfrac{P }{P (1 – P)}  dP\]

Canceling $ P$ from the denominator and numerator of the equation:

\[ \int dt = \int \dfrac{1 }{P } dP + \int \dfrac{1 }{ (1 – P)} dP\]

Solving the above equation now:

\[ t + c_1 = \ln{\left|P \right|\ -\ }\ln{\left|1-P\right|\ } \]

\[ t + c_1 =\ln{\left|\ \frac{ P }{ 1 – P }\ \right|} \]

\[ e^{ t + c_{1} } =e^{\ln{\left|\ \dfrac{ P }{ 1 – P }\ \right|}} \]

We know that $ e^{\ln{x} } = x $ so we have the above equation as:

\[ e^{ t} e^{ c_1 } = \left| \dfrac { P }{ 1 – P }  \right| \]

\[ \left| \dfrac { P }{ 1-P }  \right| = e^{ t} e^{ c_1 } \]

\[  \dfrac { P }{ 1-P }  = \pm e^{ t} e^{ c_1 } \]

Let us suppose that another constant $c $ is introduced in the equation which is $ \pm e^{ c_1 }  = c $. Now the equation becomes:

\[  \dfrac { P }{ 1-P }  = ce^{ t}  \]

Multiplying by $ 1-P $ on both the side of the equation:

\[ P=c e^t (1-P) \]

\[ P = ce^t- ce^{t}P\]

\[P+ ce^{t}P = ce^t\]

\[P(1+ ce^{t}) = ce^t\]

\[P= \dfrac{ce^t}{(1+ ce^{t})}\]

Numerical Result

\[P= \dfrac{ce^t}{(1+ ce^{t})}\]

Example

Integrate the equation:

\[\int dt= \int \dfrac{1}{x } dx \]

Solving the above equation now:

\[t+c_1 = \ln{\left|x \right|}\]

\[e^{t+ c_1}=e^{\ln{x}}\]

We know that $ e^{\ln{x}} = x $ so we have the above equation as:

\[e^{t} e^{ c_1}=x\]

5/5 - (7 votes)