Home Practice Questions AP Calculus AB Differential Equations Question No. 2

Answer is: option3

Solution:

\( \frac{dy}{dx} = 0 \) on the line \( y = x \)
\( \frac{dy}{dx} = 1 \) on the line \( y = x + 1 \)

Option (A): \( \frac{dy}{dx} = x + y \)

1. On \( y = x \): \( \frac{dy}{dx} = x + x = 2x \rightarrow \text{not zero unless } x = 0 \ ❌ \)
2. On \( y = x + 1 \): \( \frac{dy}{dx} = x + (x + 1) = 2x + 1 \rightarrow \text{not always } 1 \ ❌ \)

Fails both conditions.

Option (B): \( \frac{dy}{dx} = x - y \)

1. On \( y = x \): \( \frac{dy}{dx} = x - x = 0 \ ✅ \)
2. On \( y = x + 1 \): \( \frac{dy}{dx} = x - (x + 1) = -1 \), but you observed slope = 1 ❌

Wrong sign for slope.

Option (C): \( \frac{dy}{dx} = -x + y \)

1. On \( y = x \): \( \frac{dy}{dx} = -x + x = 0 \ ✅ \)
2. On \( y = x + 1 \): \( \frac{dy}{dx} = -x + (x + 1) = 1 \ ✅ \)

✅ Matches observed slope of 1

Option (D): \( \frac{dy}{dx} = x^2 - y \)

1. On \( y = x \): \( \frac{dy}{dx} = x^2 - x \rightarrow \text{not zero unless } x = 0 \text{ or } x = 1 \ ❌ \)
2. On \( y = x + 1 \): \( \frac{dy}{dx} = x^2 - (x + 1) \rightarrow \text{not equal to 1 in general} \ ❌ \)

Final Answer: (C) \( \frac{dy}{dx} = -x + y \)