**From the problem given bellow ****sketch the graph f(y) versus y, determine the critical points, and classify each one as asymptotically stable or unstable. Thing is, how do you get the critical points?**

**$ \dfrac{dy}{dt}=ay + by^2$**

The aim of this question is to find the **derivative** of the given expression and sketch the graphs for different points and these points show the expression is **asymptotically** stable or not.

Moreover, this question is based on the concepts of algebra. The** critical points** are those points at which the derivative is zero. The asymptote of a curve is defined as a line, i.e., the distance between the curve and the line approaches zero.

## Expert Answer:

\[ \dfrac{dy}{dt} = f(y) = ay + by^2 \]

\[ = 2y + 4y^2 \]

Thus, the graph is as follow.

To find the critical points, we put**\[ f(y) = 0 \]**

Therefore,

\[ ay + by^2 = 0 \]

\[ y(a + by) = 0 \]

Hence, the critical points are as follows.

$y = 0$ and $y = \dfrac{-a}{b}$

To find the point of inflation, we take the second derivative of the equation,

\[ \dfrac{d^2y}{dt^2} = a \dfrac{dy}{dt} + 2by \dfrac{dy}{dt} \]

\[ = (a + 2by)\dfrac{dy}{dt} \]

\[ = (a + 2by)(ay + by^2) \]

Hence, we have the following points at which the second derivative become zero.

$y = \dfrac{-a}{2b}$, $y = 0$, and $y = \dfrac{-a}{b}$

However, we know that $y = 0$ and $y = \dfrac{-a}{b}$ are the solution of the given equation. So, the **critical point** is

**$y = \dfrac{-a}{2b}$**

The graph given above give us the following information.

$y$ is increasing, when;

$\dfrac{dy}{dt} > 0$ for $y < \dfrac{-a}{b}$

$\dfrac{dy}{dt} < 0$ for $y = \dfrac{-a}{b}$, and $\dfrac{dy}{dt} > 0$ for $y > 0$

Hence, **concavity** changes at $y = \dfrac{-a}{2b}$

So, $y = 0$ is an **unstable point** and $y = \dfrac{-a}{b}$ is a **stable point**.

## Numerical Results:

The **critical points** are as follows.

$y = 0$ and $y = \dfrac{-a}{b}$

**Concavity** changes at $y = \dfrac{-a}{2b}$

$y = 0$ is an **unstable point** and $y = \dfrac{-a}{b}$ is a **stable point**.

## Example:

Solve the following differential equation.

\[ 2xy + 1 + (x^2 + 2y)y’ \]

**Solution:**

\[ 2xy + (x^2 + 2y)y’ = 2xy + x^2y’ + 2yy’ + 1 \]

\[ = \dfrac{d}{dx}(x^2y + y^2) = -1 \]

\[ = d(x^2y + y^2) = -dx \]

By **integrating** both sides, we have,

\[ x^2y + y^2 = -x + C \]

**\[ x + x^2y + y^2 = + C \]**

*Images are created by using GeoGebra.*