Sketch a phase line diagram for dydty3 2y2 the identify the...

Sketch a phase line diagram for dy/dt = y^3 - 2y^2, then identify the equilibrium solutions as stable, unstable, or semi-stable. Also, sketch the three solutions corresponding to y(0) = 3, y(0) = -1, and y(0) = 1.

Transcript

00:01 Hello students, it is given that dy by dt equal to y cube minus 2y square.

00:07 Equilibrium point can be calculated as y cube minus 2y square equal to zero which implies that y square into y minus 2 equal to zero.

00:17 Therefore, y is equal to zero and y is equal to two.

00:20 Now for y is less than zero.

00:23 Let us take y equal to minus one.

00:25 So which implies that minus one square into minus one minus two equal to plus one into minus three.

00:33 Again, that is less than zero.

00:35 Now when y is greater than zero, let us take y is equal to one here, which implies that one square into one minus two equal to minus one less than zero.

00:48 Now both are zero.

00:51 So y is equal to zero and therefore it is semi -stable.

00:55 Moving to next part, y is less than two.

01:05 So y equal to one again one square into one minus two which is equal to minus one less than zero.

01:12 Y is greater than two.

01:13 We have y equal to three which implies that three square into three minus two equal to nine into one greater than zero.

01:21 So phase diagram can be drawn as like this.

01:26 So this is zero.

01:27 This is one.

01:28 This is two minus one minus two.

01:31 So we have one point here and the phase diagram will be, it is in this direction and from here it is in this direction.

01:46 One point at zero and one point at two.

01:54 Now next, for next it is given that integral of dy divided by y square into y minus two equal to integral of dt...

Question

Sketch a phase line diagram for dy/dt = y^3 - 2y^2, then identify the equilibrium solutions as stable, unstable, or semi-stable. Also, sketch the three solutions corresponding to y(0) = 3, y(0) = -1, and y(0) = 1.

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