Question

In Problems 21–28 find the critical points and phase portrait of the given autonomous first-order differential equation. Classify each critical point as asymptotically stable, unstable, or semi-stable. By hand, sketch typical solution curves in the regions in the xy-plane determined by the graphs of the equilibrium solutions.

$\begin{array}{l}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{y}\mathbf{l}\mathbf{n}\mathbf{}\mathbf{(}\mathbf{y}\mathbf{+}\mathbf{2}\mathbf{)}\\ \end{array}$

Short answer

• The critical point 0 is unstable and the critical point -1 is asymptotically stable.
• The sketch is drawn below.

Step-by-step solution

Critical points

The zeros of the function f in $dy/dx=f\left(y\right)$are of special importance. Wesay that a real number c is a critical point of the autonomous differential equation$dy/dx=f\left(y\right)$ if it is a zero of f—that is, $f\left(c\right)=0$. A critical point is also called an equilibrium point or stationary point.

Find the critical point and phase portrait

To find the critical points, equate$\frac{dy}{dx}=0$

So,

$\begin{array}{c}y\ln (y+2)=0\\ y=0,\ln (y+2)=0\\ y=0,y+2=1\\ y=0,y=-1\end{array}$

Hence, the critical points of the given differential equations are$y=0,$and$y=-1$.

The phase portrait is shown below:

The arrows are moving away from the critical point 0. Hence, the critical point 0 is unstable.

The arrows are pointing towards the critical point -1. Hence, the critical point -1 is asymptotically stable.

Sketch the typical solution

Sketch of the family of typical solution curves of the given differential equation is shown below:

Therefore, the critical point 0 is unstable and the critical point -1is asymptotically stable.