Question

Consider the differential equation $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{y}\left(a-by\right)$wherea andb are positive constants.

(a) Either by inspection or by the method suggested in Problems 37-40, find two constant solutions of the DE.

(b) Using only the differential equation, find intervals on the y-axis on which a nonconstant solution$\mathit{y}\mathbf{=}\mathit{\varphi }\left(x\right)$is increasing. Find intervals on which $\mathit{y}\mathbf{=}\mathit{\varphi }\left(x\right)$is decreasing.

(c) Using only the differential equation, explain why $\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathbf{/}\mathbf{2}\mathit{b}$is the y-coordinate of a point of inflection of the graph of a nonconstant solution $\mathit{y}\mathbf{=}\mathit{\varphi }\left(x\right)$.

(d) On the same coordinate axes, sketch the graphs of the two constant solutions found in part (a). These constant solutions partition the$\mathit{y}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$-plane into three regions. In each region, sketch the graph of a nonconstant solution$\mathit{y}\mathbf{=}\mathit{\varphi }\left(x\right)$whose shape is suggested by the results in parts (b) and (c).

Short answer

(a) Two constant solutions: $y=0,y=a/b$

(b) Solution is increasing onand decreasing on.

(c) Inflection point:$y=\frac{a}{2b}$

(d) The graph is drawn below.

Step-by-step solution

Definition of Ordinary Differential Equation

A system of ordinary differential equations is two or more equations involving the derivatives of two or more unknown functions of a single independent variable.

Find the constant solutions

a.

Given $\frac{dy}{dx}=y(a-by)$

Remember the constant solution

$\frac{dy}{dx}=0$

Therefore, we get

$\frac{dy}{dx}=y\left(a-by\right)=0$

$y=0,a-by=0$

$by=a$

$y=a/b$

Hence, the two constant solutions of the given DE are:

$\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{=}\mathit{a}\mathbf{/}\mathit{b}$.

Find the increasing and decreasing solution

b.

First, to find interval on the y-axis on which a nonconstant solution $y=\varphi \left(x\right)$ is increasing, solve the following inequality for :

$\frac{dy}{dx}>0$

$y\left(a-by\right)>0$

$0<y<\frac{a}{b}$

On the other side, interval on the -axis on which a nonconstant solution $y=\varphi \left(x\right)$is decreasing will be:

$\frac{dy}{dx}<0$

$y\left(a-by\right)<0$

$y<0ory>\frac{a}{b}$

Hence, theSolution is increasing on$<0,\frac{a}{b}>$and decreasing on $\mathit{y}\mathbf{\in }<-\infty ,0>\mathbf{\cup }<\frac{a}{b},\infty >$.

Calculate the differential equation

c.

$$\begin{gathered} \left(d^{2}y\right)/\left(d{x}^2\right)=\frac{d}{dx}(dy/dx) \\ =\frac{d}{dx}\left[y\left(a-by\right)\right] \\ =\frac{dy}{dx}\left(a-by\right)-by\frac{dy}{dx} \\ =\frac{dy}{dx}\left(a-2by\right) \end{gathered}$$

It still remains to equate the obtained to 0.

$\frac{dy}{dx}\left(a-2by\right)=0$

$a-2by=0$

$y=\frac{a}{2b}$

Which is an inflection point.

Hence,is the y-coordinate of a point of inflection of the graph of a nonconstant solution.

Graph

d.

Hence, this is the graph of a nonconstant solution $\mathit{y}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$whose shape is suggested by the results in parts (b) and (c).