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sesenic [268]
3 years ago
6

Sketch the equilibrium solutions for the following DE and use them to determine the behavior of the solutions.

Mathematics
1 answer:
GREYUIT [131]3 years ago
4 0

Answer:

y=\dfrac{1}{1-Ke^{-t}}

Step-by-step explanation:

Given

The given equation is a differential equation

\dfrac{dy}{dt}=y-y^2

\dfrac{dy}{dt}=-(y^2-y)

By separating variable

⇒\dfrac{dy}{(y^2-y)}=-t

\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-dt

Now by taking integration both side

\int\left(\dfrac{1}{y-1}-\dfrac{1}{y}\right)dy=-\int dt

⇒\ln (y-1)-\ln y=-t+C

Where C is the constant

\ln \dfrac{y-1}{y}=-t+C

\dfrac{y-1}{y}=e^{-t+c}

\dfrac{y-1}{y}=Ke^{-t}

y=\dfrac{1}{1-Ke^{-t}}

from above equation we can say that

When t  will increases in positive direction then e^{-t} will decreases it means that {1-Ke^{-t}} will increases, so y will decreases. Similarly in the case of negative t.

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