▶️ Answer/Explanation
Solution
The slope field shows slopes depending on both \( x \) and \( y \).
(A) \(\frac{dy}{dx} = 1 + x\): Depends only on \( x \), slopes constant for fixed \( x \).
(B) \(\frac{dy}{dx} = x^2\): Depends only on \( x \), slopes zero at \( x = 0 \).
(C) \(\frac{dy}{dx} = x + y\): Slope zero along \( y = -x \), positive above, negative below.
(D) \(\frac{dy}{dx} = \frac{x}{y}\): Slope zero at \( x = 0 \), positive for \( x > 0 \), \( y > 0 \).
(E) \(\frac{dy}{dx} = \ln y\): Depends only on \( y \), slopes constant for fixed \( y \).
The field’s diagonal pattern with slopes transitioning across \( y = -x \) matches (C).
✅ Answer: C