Question

Find the critical points and sketch the phase portrait of the given autonomous first-order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable.

y′=-y²+5y+6

Answer 1

To find the critical points and sketch the phase portrait of the given differential equation y′=-y²+5y+6, we need to find where the derivative is zero. We can use the second derivative test to determine the stability of the critical points and sketch the phase portrait.

Step-by-step explanation:

To find the critical points and sketch the phase portrait of the given differential equation y′=-y²+5y+6, we need to find where the derivative is zero. Setting the derivative equal to zero, we have:

0 = -y²+5y+6

Factoring the equation, we get:

(-y+2)(y+3) = 0

This gives us two critical points: y = 2 and y = -3.

To determine the stability of these critical points, we can use the second derivative test. The second derivative of the given differential equation with respect to y is:

y'' = 2y - 5

We can evaluate the second derivative at the critical points:

y''(2) = 2(2) - 5 = -1

y''(-3) = 2(-3) - 5 = -11

Since y''(2) = -1 is negative, the critical point y = 2 is an asymptotically stable critical point. And since y''(-3) = -11 is also negative, the critical point y = -3 is an asymptotically stable critical point as well.

Based on these stability properties, we can sketch the phase portrait:

Phase Portrait:

At the critical point y = 2, the system is attracted towards this point, and similarly, at the critical point y = -3, the system is attracted towards this point. Therefore, the phase portrait will have arrows pointing towards y = 2 and y = -3.