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.
Figure 1: A graph between f(y) and y
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.