Question

Consider the differential equation$\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}\mathbf{=}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$

(a) Explain why a solution of the DE must be an increasing function on any interval of the x-axis.

(b) What are and $\underset{\mathbf{x}\mathbf{\to }\mathbf{\infty }}{\mathbf{l}\mathbf{i}\mathbf{m}}\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{x}$? What does this suggest about a solution curve as $\mathit{x}\mathbf{\to }\mathbf{\pm }\mathbf{\infty }$?

(c) Determine an interval over which a solution curve is concave down and an interval over which the curve is concave up.

(d) Sketch the graph of a solution $\mathit{y}\mathbf{=}\mathit{\varphi }\mathbf{\left(}\mathit{x}\mathbf{\right)}$of the differential equation whose shape is suggested by parts (a)–(c).

Short answer

a. Because $e^{-x^2}>0$, for all x.

b. $\underset{x\to -\infty }{\lim }=\underset{x\to +\infty }{\lim }=0$, the solution has two horizontal asymptotes.

c. The solution is concave up on $x\in (-\infty ,0)$and concave down on $x\in (0,\infty )$.

Step-by-step solution

Definition of differential equation.

A differential equation is defined as the derivative function of one or more unknown functions (dependable variables) with respect to one or more undependable variables.

Solution for part a)

a.

Let $f$be a function and let $D$be its domain. If $f\text{'}\left(x\right)>0$for all values $x\in D$, then is an increasing function.

Notice that $\frac{dy}{dx}$is always greater than zero because:

$e^{-x^2}>0$, for all

Since the first derivative of y is always greater than zero, it means that the solution y is increasing.

Solution for part b)

If,

$\underset{x\to -\infty }{\lim }\frac{dy}{dx}=\underset{x\to -\infty }{\lim }{e}^{-x^2}$

Then,

$\underset{x\to -\infty }{\lim }\frac{dy}{dx}=0$

If,

$\underset{x\to +\infty }{\lim }\frac{dy}{dx}=\underset{x\to +\infty }{\lim }{e}^{-x^2}$

Then,

$\underset{x\to +\infty }{\lim }\frac{dy}{dx}=0$

The limit suggest the slope of tangents in points when x increases or decreases approaches zero(a horizontal line has slope zero), therefore, the solution has two horizontal asymptotes.

Solution for part c)

Find the derivative of

$$\begin{gathered} \frac{dy}{dx}=e^{-x^2} \\ \frac{d^{2}y}{d{x}^2}=-2x{e}^{-x^2} \end{gathered}$$

The points of inflection are when $\frac{d^{2}y}{d{x}^2}=0$, therefore,

$-2x{e}^{-x^2}\iff x=0$

Note that when $x\in (-\infty ,0)$, then $\frac{d^{2}y}{d{x}^2}=-2x{e}^{-x^2}>0$and

When $x\in (0,\infty )$, then $\frac{d^{2}y}{d{x}^2}=-2x{e}^{-x^2}<0$.

Therefore, the solution is concave up on $x\in (-\infty ,0)$and concave down on $x\in (0,\infty )$.

Solution for part d)

The sketch is shown below.

Note that we have an increasing solution with two horizontal asymptotes and the function is concave up for $x\in (-\infty ,0)$and concave down for $x\in (-\infty ,0)$.

The shown sketch is a part of the direction field of the given differential equation.

Therefore,

1. Because $e^{-x^2}>0$, for all x.
2. $\underset{x\to -\infty }{\lim }=\underset{x\to +\infty }{\lim }=0$, the solution has two horizontal asymptotes.
3. The solution is concave up on $x\in (-\infty ,0)$and concave down on $x\in (-\infty ,0)$.