Slope Fields for 5 DEs: Tables & Analysis

Answer

A slope field (or direction field) is a graphical representation of the solutions to a first-order differential equation, where small line segments are drawn at grid points $(x, y)$ with a slope equal to the value of $\frac{dy}{dx}$ at that point. Based on the provided image, we will analyze the behaviors of five specific differential equations on a $6 \times 5$ grid spanning $x \in [-2, 3]$ and $y \in [-2, 2]$.

Image Observation

The image displays a Cartesian coordinate system with a grid of dots. There are 6 dots horizontally and 5 dots vertically. The $x$-axis shows points from $-2$ to $3$, and the $y$-axis shows points from $-2$ to $2$. This gives us 30 specific coordinate pairs where slope segments must be drawn.

Explanation

1. Analysis of Question 1: $\frac{dy}{dx} = \frac{x}{y}$ The slope is determined by the ratio of the $x$-coordinate to the $y$-coordinate.

   $y \setminus x$	-2	-1	0	1	2	3
   2	-1	-0.5	0	0.5	1	1.5
   1	-2	-1	0	1	2	3
   0	und	und	und	und	und	und
   -1	2	1	0	-1	-2	-3
   -2	1	0.5	0	-0.5	-1	-1.5

   $\frac{dy}{dx} = \frac{x}{y}$ This formula indicates that slopes are zero along the y-axis ($x=0$) and undefined (vertical) along the x-axis ($y=0$). ⚠️ This step is required on exams: Identify where the derivative is zero or undefined first to anchor your sketch.

2. Analysis of Question 2: $\frac{dy}{dx} = -3x$ This slope depends only on the $x$ value. Therefore, all segments in a vertical column will have the same steepness.

   $x$	-2	-1	0	1	2	3
   Slope	6	3	0	-3	-6	-9

   $\frac{dy}{dx} = -3x$ This linear relationship shows that as $x$ increases, the slope becomes increasingly negative and steep.

3. Analysis of Question 3: $\frac{dy}{dx} = y + 1$ This slope depends only on the $y$ value. All segments in a horizontal row will be identical.

   $y$	2	1	0	-1	-2
   Slope	3	2	1	0	-1

   $\frac{dy}{dx} = y + 1$ The slopes are zero along the horizontal line $y = -1$, representing an equilibrium solution.

4. Analysis of Question 4: $\frac{dy}{dx} = x - 2$ Similar to Question 2, the slope is constant for each vertical column and depends only on the distance from $x=2$.

   $x$	-2	-1	0	1	2	3
   Slope	-4	-3	-2	-1	0	1

   $\frac{dy}{dx} = x - 2$ The slope is zero at $x=2$ and becomes positive only when $x > 2$.

5. Analysis of Question 5: $\frac{dy}{dx} = y - x$ This requires calculating the difference for every point. The slope is zero whenever $y = x$.

   $y \setminus x$	-2	-1	0	1	2	3
   2	4	3	2	1	0	-1
   1	3	2	1	0	-1	-2
   0	2	1	0	-1	-2	-3
   -1	1	0	-1	-2	-3	-4
   -2	0	-1	-2	-3	-4	-5

   $\frac{dy}{dx} = y - x$ The slope values follow a diagonal pattern where segments are parallel along lines $y = x + C$.

Final Answer

The slope fields are constructed by drawing small lines with the calculated slopes at each of the 30 coordinates. $\boxed{\text{See tables for slope values at each point}}$

Common Mistakes

• Confusing zero vs. undefined: In Question 1, when $y=0$, the slope is undefined; draw a vertical line. When $x=0$, the slope is zero; draw a horizontal line.
• Uniformity: Forgetting that if an equation only contains $x$ (like Q2 and Q4), the slopes in any vertical column must be identical. If it contains only $y$ (like Q3), the slopes in any horizontal row must be identical.