Chapter 7.1, Problem 72E

To determine

To find the correct differential equation from the given choices for which the slope field show only negative slopes in the fourth quadrant.

Answer to Problem 72E

The slope field which shows only negative slopes in the fourth quadrant from the given choices is $\frac{dy}{dx}=\frac{y}{x^2}-3$ .

Explanation of Solution

Given information :

The choices of differential equations are,

1. $\frac{dy}{dx}=-\frac{x}{y}$
2. $\frac{dy}{dx}=xy+5$
3. $\frac{dy}{dx}=x{y}^2-2$
4. $\frac{dy}{dx}=\frac{x^3}{x^2}$
5. $\frac{dy}{dx}=\frac{y}{x^2}-3$

Calculation:

It is clear that on the fourth quadrant of the slope field x -coordinates are positive and y -coordinates are negative.

Consider, option A $\frac{dy}{dx}=-\frac{x}{y}$

  $\frac{dy}{dx}=-\frac{x}{y}\Rightarrow -\frac{\text{Positive}}{\text{Negative}}=\text{Positive}$

Hence, option A is eliminated (Since the slope field of the differential equation $\frac{dy}{dx}=-\frac{x}{y}$ contain positive slopes).

Now, consider option B $\frac{dy}{dx}=xy+5$ also take a point $\left(1,-2\right)$ from the fourth quadrant,

  $\Rightarrow \frac{dy}{dx}=xy+5\Rightarrow 1\left(-2\right)+5=-2+5=3$

Here the differential equation $\frac{dy}{dx}=xy+5$ is not negative for all the slopes.

Hence, option B is eliminated (Since the slope field of the differential equation $\frac{dy}{dx}=xy+5$ does not contain all negative slopes).

Consider option C $\frac{dy}{dx}=x{y}^2-2$ again consider a point $\left(1,-2\right)$ from the fourth quadrant,

  $\frac{dy}{dx}=x{y}^2-2\Rightarrow 1{\left(-2\right)}^2-2=4-2=2$

Here the differential equation $\frac{dy}{dx}=x{y}^2-2$ is not negative for all the slopes.

Hence, option C is eliminated (Since the slope field of the differential equation $\frac{dy}{dx}=x{y}^2-2$ does not contain all negative slopes).

Now, consider option D $\frac{dy}{dx}=\frac{x^3}{x^2}$ ,

  $\frac{dy}{dx}=\frac{x^3}{x^2}\Rightarrow -\frac{{\left(\text{Positive}\right)}^3}{{\left(\text{Positive}\right)}^2}=\text{Positive}$

Hence, option D is eliminated (Since the slope field of the differential equation $\frac{dy}{dx}=\frac{x^3}{x^2}$ contain positive slopes).

Consider option E $\frac{dy}{dx}=\frac{y}{x^2}-3$ ,

It is evident that for all the values of x and y the slope of $\frac{dy}{dx}=\frac{y}{x^2}-3$ is negative.

Hence, the slope field which shows only negative slopes in the fourth quadrant from the given choices is $\frac{dy}{dx}=\frac{y}{x^2}-3$ .