Question

Multiple Choice For which of the following differential equa- tions will a slope field show nothing but negative slopes in the fourth quadrant? (A) \(\frac{d y}{d x}=-\frac{x}{y} \quad\) (B) \(\frac{d y}{d x}-x y+5 \quad\) (C) \(\frac{d y}{d x}=x y^{2}-2\) (D) \(\frac{d y}{d x}=\frac{x^{3}}{y^{2}} \quad\) (E) \(\frac{d y}{d x}=\frac{y}{x^{2}}-3\)

Short answer

Options (A), (B), (C) and (E) will show nothing but negative slopes in the fourth quadrant.

Step-by-step solution

Analyzing option (A) \(\frac{dy}{dx} = - \frac{x}{y}\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = - \frac{-a}{-b} = -\frac{a}{b}< 0\). So yes, the slopes in the 4th quadrant are negative for option A.

Analyzing option (B) \(\frac{dy}{dx} = xy - 5\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = -ab - 5 < 0\). So yes, the slopes in the 4th quadrant are negative for option B.

Analyzing option (C) \(\frac{dy}{dx} = xy^2 - 2\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = -ab^2 - 2 < 0\). So yes, the slopes in the 4th quadrant are negative for option C.

Analyzing option (D) \(\frac{dy}{dx} = \frac{x^3}{y^2}\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = \frac{(-a)^3}{(-b)^2} = -a^3/b^2 > 0\). So no, the slopes in the 4th quadrant are not negative for option D.

Analyzing option (E) \(\frac{dy}{dx} = \frac{y}{x^2} - 3\)

Replacing \(x\) and \(y\) with their fourth quadrant values: \(x = -a, y = -b\) where \(a, b > 0\).Substituting these into the equation: \(\frac{dy}{dx} = \frac{-b}{(-a)^2} - 3 = -\frac{b}{a^2} - 3 < 0\). So yes, the slopes in the 4th quadrant are negative for option E.