Solve the given differential equation by separation of variables. dPdt=P−P2

Name: Solve the given differential equation by separation of variables. dPdt=P−P2
Uploaded: 2022-02-23T17:22:57+00:00
Description: Solve the given differential equation by separation of variables. dPdt=P−P2

Question

Solve the given differential equation by separation of variables.  dPdt=P−P2

AGRFTr · Accepted Answer

Solving the separable DE, first we multiply the whole equation by $\frac{d t}{P - P^{2}}$ to get $\frac{d P}{P - P^{2}} = d t \to \int \frac{d P}{P - P^{2}} = \int d t$
$∴ \int \frac{d P}{P (1 - P)} = \int d t$
Partial fractions can be used to solve the L.H.S integral, and we will format them as follows:
$\frac{A}{P} + \frac{B}{P - 1} = \frac{1}{P (P - 1)}$ , where A and B are constants
We can determine the values of A and B using the cover up rule
First, cover up rule, we can get the values of A and B
First, cover up P and substitute by P=0 to get A
$∴ \frac{1}{1 - 0} = 1 \to A = 1$
Then, cover up (1-P) and substitute by P=1 to get B
$∴ \frac{1}{1} = 1 \to B = 1$
Reintroduce substitution into the equation to obtain
$\int (\frac{1}{P} + \frac{1}{1 - P}) d p = \int d t \to \ln | P | - \ln | 1 - P | = t + c_{1}$
$∴ \ln | \frac{P}{1 - P} | = t + c_{1} =$
Exponentiating both sides, we get
$\frac{P}{1 - P} = e^{t + c_{1}} = e^{t} c^{c_{1}} = C e^{t}$
$∴ P = (1 - P) C e^{t} \to P + P C e^{t} = C e^{t}$
$∴ P (1 + C e^{t}) = C e^{t} \to P = \frac{C e^{t}}{1 + C e^{t}}$
Result:
$P = \frac{C e^{t}}{1 + C e^{t}}$

Solve the given differential equation by separation of variables. \frac

Answered question

Answer & Explanation

New Questions in Differential Equations