This calculus exercise asks students to match four differential equations with initial conditions to their corresponding slope fields labeled (a) through (d).

To solve this matching problem, we need to analyze each differential equation and understand how its slope field should behave based on the formula for $\frac{dy}{dx}$. Then we match it to one of the four slope fields (a), (b), (c), (d) by comparing key features such as:

- Where slopes are zero (horizontal segments)
- Symmetry
- Behavior as $x$ or $y$ becomes large
- Specific values at given points (especially initial conditions)

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## Equation (1): $\frac{dy}{dx} = 1 - yx$, with $y(1) = 1$

Analysis:

- Slope is zero when $1 - yx = 0 \Rightarrow y = \frac{1}{x}$ → hyperbola.
- At point $(1,1)$: slope = $1 - (1)(1) = 0$ → horizontal tangent.
- For $x > 0$, if $y > 1/x$, slope is negative; if $y < 1/x$, slope is positive.
- For $x < 0$, signs reverse.

Look at slope field (d):
→ It has a curve-like region where slopes are horizontal — likely along $y = 1/x$.
→ At (1,1), there’s a horizontal segment.
→ The slope field shows symmetry consistent with $yx = 1$ being the zero-slope curve.

✔ Match: (1) → (d)

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## Equation (2): $\frac{dy}{dx} = \frac{x}{1 - y}$, with $y(0) = -3$

Analysis:

- Slope is undefined when $y = 1$ → vertical asymptote in slope field (vertical segments or gaps).
- Slope is zero when $x = 0$ → along the y-axis.
- At (0, -3): slope = 0 → horizontal tangent.
- For $y < 1$, sign of slope follows sign of $x$: right side ($x>0$) → positive slope; left side ($x<0$) → negative slope.

Look at slope field (c):
→ Has vertical lines near $y=1$ (undefined slope).
→ Along x=0, slopes are horizontal → matches zero slope.
→ For $x>0$, slopes point up-right; for $x<0$, down-left → matches sign of $x$.

✔ Match: (2) → (c)

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## Equation (3): $\frac{dy}{dx} = x^2 + y^2 - 1$, with $y(1) = 0$

Analysis:

- Slope is zero when $x^2 + y^2 = 1$ → unit circle.
- Inside the unit circle ($x^2 + y^2 < 1$), slope is negative.
- Outside ($x^2 + y^2 > 1$), slope is positive.
- At (1,0): $x^2 + y^2 = 1$ → slope = 0 → horizontal tangent.

Look at slope field (a):
→ Clearly shows a circular region (unit circle) where slopes are horizontal.
→ Inside the circle, slopes point downward (negative); outside, upward (positive).
→ Perfect match for $x^2 + y^2 - 1$.

✔ Match: (3) → (a)

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## Equation (4): $\frac{dy}{dx} = y - \sin(x)$, with $y(0) = 1$

Analysis:

- Slope is zero when $y = \sin(x)$ → sine wave.
- At (0,1): slope = $1 - \sin(0) = 1 - 0 = 1$ → positive slope.
- The slope field should oscillate vertically following the sine curve.

Look at slope field (b):
→ Shows wavy horizontal regions — consistent with $y = \sin(x)$.
→ At (0,1), which is above the sine curve (since sin(0)=0), slope should be positive → indeed, arrows point upward.
→ The slope field varies periodically with x, as expected from $\sin(x)$.

✔ Match: (4) → (b)

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## ✔ Final Matching:

| Differential Equation | Slope Field |
|------------------------|-------------|
| (1) $\frac{dy}{dx} = 1 - yx$; $y(1)=1$ | (d) |
| (2) $\frac{dy}{dx} = \frac{x}{1 - y}$; $y(0)=-3$ | (c) |
| (3) $\frac{dy}{dx} = x^2 + y^2 - 1$; $y(1)=0$ | (a) |
| (4) $\frac{dy}{dx} = y - \sin(x)$; $y(0)=1$ | (b) |

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## 🖋️ Sketching the Solution Curves:

For each, sketch a curve starting at the initial condition that follows the direction of the slope field:

- (1) → (d): Start at (1,1), follow horizontal slope, then curve according to field — likely decreasing initially since for x>1, y=1 gives negative slope (if x increases beyond 1, slope becomes negative).
- (2) → (c): Start at (0,-3), horizontal, then move right/up since for x>0, slope positive; avoid y=1 (asymptote).
- (3) → (a): Start at (1,0), horizontal, then since outside unit circle, slope positive → curve goes up and right.
- (4) → (b): Start at (0,1), slope = 1 → go up-right, then follow the wave pattern — will oscillate around y=sin(x).

*(Note: Since you can’t draw here, describe the path as following the slope arrows from the initial point.)*

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## ✔ Final Answer:

(1) → (d)
(2) → (c)
(3) → (a)
(4) → (b)