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.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{y}{\mathbf{e}}^{\mathbf{y}}\mathbf{-}\mathbf{9}\mathbf{y}}{{\mathbf{e}}^{\mathbf{y}}}$

Short answer

The critical point 2 In 3 is unstable and the critical point 0 is stable.

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,

localid="1668424239737" $$\begin{gathered} \frac{y{e}^y-9y}{e^y}=0 \\ y{e}^y-9y=0 \\ y=0,e^y-9=0 \\ y=0,y=2\ln \end{gathered}$$

Hence, the critical points of the given differential equations are localid="1668424242825" $y=0,$and localid="1668424245937" $y=2\ln 3$.

The phase portrait is shown below:

Since both arrows are pointing away from 2 In 3, so it is unstable critical point.

Since both arrows are pointing towards 0, it is a stable critical point.

Step 3:Sketch the typical solution.

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

Therefore, the critical point 2 In 3 is unstable and the critical point 0 is stable.