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Differential Equations - Bernoulli equation

ajkess1994

New member
Nov 10, 2018
9
Afternoon, anyone that would like to take a look at this Differential Equation problem it would be very helpful. I have tried separating the problem, but I am only working with one known term.

Consider the logistic equation

$$\dot{y}=y(1-y). $$

(a) Find the solution satisfying $y_1(0)=6$ and $y_2(0)=−1. $

$y_1(t)= ?$

$y_2(t)= ?$

(b) Find the time $t$ when $y_1(t)=3. $

$t= ?$

(c) When does $y_2(t)$ become infinite?

$t= ?$
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
Afternoon, anyone that would like to take a look at this Differential Equation problem it would be very helpful. I have tried separating the problem, but I am only working with one known term.

Consider the logistic equation

[y˙=y(1−y)] NOTICE: There is a small dot above the first y term, NOT off to the side

(a) Find the solution satisfying y1(0)=6 and y2(0)=−1.

y1(t)= ?

y2(t)= ?

(b) Find the time t when [y1(t)=3].

t= ?

(c) When does y2(t) become infinite?

t= ?
To get you started: This is a Bernoulli differential equation. You can see the solution method here.

-Dan
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
Or another approach:
\begin{align*}
\dot{y}&=y(1-y) \\
\int\frac{dy}{y(1-y)}&=\int dt \\
\int\frac{dy}{y}-\int\frac{dy}{y-1}&=t+C \\
\ln|y|-\ln|y-1|&=t+C \\
\ln\left|\frac{y}{y-1}\right|&=t+C \\
\frac{y}{y-1}&=Ce^t \\
&\vdots
\end{align*}
You can solve for $y$ from here.
 

ajkess1994

New member
Nov 10, 2018
9
Thank you Dan

- - - Updated - - -

Thank you Ackbach