Question

select the differential equation that matches the given slope field.
answer
\\(\frac{dy}{dx} = -\frac{x^2}{y-1}\\) \\(\frac{dy}{dx} = x(y - 1)\\)
\\(\frac{dy}{dx} = \frac{x}{(y-1)^2}\\) \\(\frac{dy}{dx} = -\frac{(y-1)^2}{x^2}\\)

Explanation:

Step1: Analyze the slope field at \( y = 1 \)

When \( y = 1 \), the slope \( \frac{dy}{dx} \) should be horizontal (slope = 0) or have a specific behavior. Let's check each option:
- For \( \frac{dy}{dx}=-\frac{x^2}{y - 1} \), when \( y = 1 \), it's undefined (division by zero). Eliminate this.
- For \( \frac{dy}{dx}=x(y - 1) \), when \( y = 1 \), \( \frac{dy}{dx}=0 \). This is a candidate.
- For \( \frac{dy}{dx}=\frac{x}{(y - 1)^2} \), when \( y = 1 \), it's undefined. Eliminate this.
- For \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \), when \( y = 1 \), \( \frac{dy}{dx}=0 \). But let's check other features.

Step2: Analyze the slope field at \( x = 0 \)

When \( x = 0 \), the slope \( \frac{dy}{dx} \) should be zero (horizontal lines) because at \( x = 0 \), the slope field has horizontal segments.
- For \( \frac{dy}{dx}=x(y - 1) \), when \( x = 0 \), \( \frac{dy}{dx}=0 \). This matches.
- For \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \), when \( x = 0 \), it's undefined (division by zero). Eliminate this.

Step3: Analyze the sign of the slope

- When \( x>0 \) and \( y>1 \), the slope should be positive (lines are increasing). For \( \frac{dy}{dx}=x(y - 1) \), \( x>0 \) and \( y - 1>0 \) implies \( \frac{dy}{dx}>0 \), which matches.
- When \( x>0 \) and \( y<1 \), \( y - 1<0 \) and \( x>0 \) implies \( \frac{dy}{dx}<0 \), but wait, in the slope field, when \( x>0 \) and \( y<1 \) (e.g., \( y = 0 \), \( x = 2 \)), the slope is positive? Wait, no, let's re - check. Wait, maybe I made a mistake. Wait, when \( x>0 \) and \( y<1 \), \( y - 1\) is negative, \( x\) is positive, so \( \frac{dy}{dx}=x(y - 1)\) is negative. But in the slope field, when \( x>0 \) and \( y<1 \) (e.g., \( x = 2 \), \( y = 0 \)), the slope seems positive? Wait, no, maybe I misread the slope field. Wait, let's check another point. When \( x<0 \) and \( y>1 \), \( x\) is negative, \( y - 1\) is positive, so \( \frac{dy}{dx}=x(y - 1)\) is negative, which matches the slope field (left of \( y\) - axis, above \( y = 1\), slopes are negative). When \( x<0 \) and \( y<1 \), \( x\) is negative, \( y - 1\) is negative, so \( \frac{dy}{dx}=x(y - 1)\) is positive, which matches the slope field (left of \( y\) - axis, below \( y = 1\), slopes are positive). When \( x>0 \) and \( y>1 \), positive slope (matches), when \( x>0 \) and \( y<1 \), negative slope? Wait, no, in the slope field, when \( x>0 \) and \( y<1 \) (e.g., \( x = 2 \), \( y = 0 \)), the slope is positive? Wait, maybe the slope field's right - hand side ( \( x>0 \)) below \( y = 1 \) has positive slopes? Wait, no, looking at the slope field, for \( x>0 \) and \( y<1 \) (e.g., \( x = 2 \), \( y = 0 \)), the slope segments are increasing (positive slope). Wait, but according to \( \frac{dy}{dx}=x(y - 1) \), when \( x>0 \) and \( y<1 \), \( y - 1<0 \), so \( \frac{dy}{dx}<0 \). There is a contradiction here? Wait, maybe I made a mistake in the slope field analysis. Wait, let's check the other option \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \). When \( x eq0 \), \( (y - 1)^2\geq0 \) and \( x^2>0 \), so \( \frac{dy}{dx}\leq0 \). But in the slope field, when \( x>0 \) and \( y>1 \), the slopes are positive, so this can't be. Wait, maybe the first analysis of \( y = 1 \) was wrong. Wait, the slope field at \( y = 1 \) has vertical lines? No, the slope field at \( y = 1 \) has horizontal lines. Wait, when \( y = 1 \), for \( \frac{dy}{dx}=x(y - 1) \), it's zero, so horizontal lines, which matches. For \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \), when \( y = 1 \), it's zero, but when \( x = 0 \), it's undefined. The slope field at \( x = 0 \) has horizontal lines, so \( x = 0 \) should have \( \frac{dy}{dx}=0 \). For \( \frac{dy}{dx}=x(y - 1) \), at \( x = 0 \), it's zero, which matches. For the other options, when \( x = 0 \):
- \( \frac{dy}{dx}=-\frac{x^2}{y - 1}=0 \) (wait, earlier I thought it was undefined, but \( x = 0 \), so \( -\frac{0}{y - 1}=0 \). Wait, I made a mistake here. Let's re - evaluate.
- For \( \frac{dy}{dx}=-\frac{x^2}{y - 1} \), when \( x = 0 \), \( \frac{dy}{dx}=0 \). When \( y = 1 \), it's undefined (division by zero). But in the slope field, at \( y = 1 \), the lines are vertical? No, the slope field at \( y = 1 \) has horizontal lines. So when \( y = 1 \), the slope should be zero (horizontal), but \( \frac{dy}{dx}=-\frac{x^2}{y - 1} \) is undefined at \( y = 1 \), so it can't be.
- For \( \frac{dy}{dx}=\frac{x}{(y - 1)^2} \), when \( x = 0 \), \( \frac{dy}{dx}=0 \), but when \( y = 1 \), it's undefined.
- Back to \( \frac{dy}{dx}=x(y - 1) \): when \( x>0 \) and \( y>1 \), slope positive (matches slope field, right of \( y\) - axis, above \( y = 1\), slopes are positive). When \( x>0 \) and \( y<1 \), \( y - 1<0 \), \( x>0 \), so slope negative. But in the slope field, when \( x>0 \) and \( y<1 \) (e.g., \( x = 2 \), \( y = 0 \)), the slope segments are increasing (positive slope). Wait, maybe the slope field's right - hand side below \( y = 1 \) has negative slopes? Maybe I misread the slope field. Let's look at the slope field again: for \( x>0 \) (right of \( y\) - axis) and \( y<1 \) (below \( y = 1\)), the slope segments are going from bottom - left to top - right? No, wait, the slope field: for \( x>0 \), \( y<1 \), the lines are like /, which is positive slope. But according to \( \frac{dy}{dx}=x(y - 1) \), it's negative. There is a problem. Wait, maybe the correct option is \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \) is wrong. Wait, let's check the sign of \( \frac{dy}{dx}=x(y - 1) \) again. When \( x>0 \), \( y>1 \): positive slope (matches, right of \( y\) - axis, above \( y = 1\), slopes are positive). When \( x>0 \), \( y<1 \): negative slope (but in the slope field, right of \( y\) - axis, below \( y = 1\), slopes are positive? No, maybe the slope field's right - hand side below \( y = 1 \) has negative slopes? Wait, the slope field: for \( x>0 \), \( y<1 \), the lines are \, which is negative slope. Oh! I made a mistake earlier. The slope segments for \( x>0 \), \( y<1 \) are \, which is negative slope. So \( \frac{dy}{dx}=x(y - 1) \) when \( x>0 \), \( y<1 \) is negative (since \( y - 1<0 \)), which matches the slope field ( \, negative slope). When \( x<0 \), \( y>1 \): \( x<0 \), \( y - 1>0 \), so \( \frac{dy}{dx}<0 \), which matches the slope field (left of \( y\) - axis, above \( y = 1\), slopes are \, negative slope). When \( x<0 \), \( y<1 \): \( x<0 \), \( y - 1<0 \), so \( \frac{dy}{dx}>0 \), which matches the slope field (left of \( y\) - axis, below \( y = 1\), slopes are /, positive slope). And at \( x = 0 \), slope is zero (horizontal lines), which matches. At \( y = 1 \), slope is zero (horizontal lines), which matches. The other options:
- \( \frac{dy}{dx}=-\frac{x^2}{y - 1} \): at \( y>1 \), \( x eq0 \), slope is negative. But in the slope field, right of \( y\) - axis, above \( y = 1\), slope is positive. So eliminate.
- \( \frac{dy}{dx}=\frac{x}{(y - 1)^2} \): at \( x>0 \), \( y>1 \), slope is positive (matches), but at \( x<0 \), \( y>1 \), slope is negative (matches), but at \( y = 1 \), it's undefined (slope field at \( y = 1 \) has horizontal lines, not vertical), so eliminate.
- \( \frac{dy}{dx}=-\frac{(y - 1)^2}{x^2} \): slope is always non - positive, but in the slope field, left of \( y\) - axis, below \( y = 1\), slope is positive. So eliminate.

Answer:

\( \boldsymbol{\frac{dy}{dx}=x(y - 1)} \) (the option with \( \frac{dy}{dx}=x(y - 1) \))