This problem tests finding critical points in slope fields. Setting dy/dx = 0 gives us x² - y² = 0, or x² = y². Among the given points, only (1,1) satisfies this equation since 1² = 1². The point (1,0) seems plausible but gives slope = 1² - 0² = 1 ≠ 0. To locate horizontal tangents in slope fields, solve the equation obtained by setting the differential equation equal to zero.

This question analyzes solution behavior for separable equations in slope fields. The differential equation dy/dx = y/x can be rewritten as dy/y = dx/x, which integrates to ln|y| = ln|x| + C. This simplifies to y = kx for some constant k, meaning all solutions are lines through the origin. The decreasing option (D) fails because these linear solutions have positive slope k in quadrant I. For equations of the form dy/dx = g(y)/h(x), separation of variables often reveals the solution structure.

This problem tests sign analysis in two-variable slope fields. At point (2,2), we calculate dy/dx = 2(1-2) = 2(-1) = -2 < 0, confirming negative slope. At (-2,2), we get dy/dx = -2(1-2) = -2(-1) = 2 > 0, showing positive slope instead. The pattern dy/dx = x(1-y) gives negative slopes when x and (1-y) have opposite signs. To determine slope signs efficiently, factor the differential equation and analyze the sign of each factor at the given point.

This problem examines long-term behavior using slope field analysis. The differential equation dy/dx = 1 - y has equilibrium at y = 1 (where dy/dx = 0). For y(0) = 3 > 1, we have dy/dx = 1 - 3 = -2 < 0, so the solution decreases toward the equilibrium y = 1. The unbounded growth option (A) fails because dy/dx < 0 whenever y > 1, preventing increase. For autonomous equations dy/dx = f(y), stable equilibria attract nearby solutions.

This question examines how slope field patterns determine solution relationships. Since dy/dx = sin x depends only on x (not on y), all points with the same x-coordinate have identical slopes. This means solution curves maintain the same vertical spacing everywhere, making them vertical translations of each other. The horizontal translation option (B) fails because sin x has different values at different x-coordinates. For differential equations of the form dy/dx = f(x), all solutions differ by only a vertical shift.

This question examines global properties of slope fields. Since 1 + y² ≥ 1 for all real y, we have 0 < dy/dx ≤ 1 everywhere, meaning all solutions have positive slope and are increasing functions. The slope equals 1 when y = 0 and approaches 0 as |y| → ∞. The negative slope option (B) fails because 1/(1+y²) cannot be negative. For rational functions in slope fields, analyze the sign and bounds of the expression to determine universal solution properties.

This question tests reasoning about solution behavior using slope fields. At the point (0,1), we calculate the slope: dy/dx = 0 - 1 = -1. Since the slope is negative, the solution curve must be decreasing as it passes through this point. The horizontal tangent option (A) would require slope = 0, which contradicts our calculation. To analyze slope fields systematically, always substitute the given point coordinates into the differential equation to determine the exact slope value.

Reasoning using slope fields involves determining regions where slopes are positive, negative, or zero to understand solution monotonicity. For dy/dx = x + y, slopes are negative precisely when x + y < 0, or y < -x. In this region below the line y = -x, solution curves decrease as they follow negative tangents. Above the line, positive slopes indicate increasing behavior. A tempting distractor is points with y > -x, but this fails as slopes are positive there, not negative. A transferable strategy for slope fields is to plot the zero-slope isocline to divide the plane into regions of consistent slope sign.

This question tests reasoning using slope fields by determining the sign of the slope at a specific point. In the slope field for dy/dx = y/(1 + x²), the denominator 1 + x² is always positive, so the sign of the slope matches the sign of y. At (1,-2), y = -2 < 0, so the slope is negative, indicating the solution curve is decreasing there. This field's slopes are positive above the x-axis and negative below, scaled by the positive denominator. A tempting distractor is positive, which would be true if y were positive, but here y is negative. A transferable slope-field strategy is to factor the differential equation to determine the sign based on the signs of numerator and denominator at the point.

This question tests reasoning using slope fields by describing the behavior of solutions in a specific region. In the slope field for $\frac{dy}{dx} = y(2 - y)$, for $0 < y < 2$, both $y > 0$ and $(2 - y) > 0$, so the product is positive, meaning all slopes are positive and solutions are increasing for all $x$. As $y$ approaches 2 from below, the slope approaches 0, so solutions increase toward the equilibrium at y = 2 asymptotically. Below $y = 0$ or above $y = 2$, slopes are negative, leading to different behaviors like decreasing toward equilibria. A tempting distractor is that they decrease for all $x$, which might occur if one flips the sign of $(2 - y)$, but it's positive in this interval. A transferable slope-field strategy is to analyze the sign of $\frac{dy}{dx}$ in different regions divided by equilibria to predict whether solutions increase or decrease.