Question

The equation of the curve whose slope is \(\frac{y-1}{x^2+x}\) and which passes through the point \((1,0)\) is

• A. \(xy-x-y-1 = 0\)
• B. \((y-1)(x+1) = 2x\)
• C. \(xy+x+y-1 = 0\)
• D. \(y(x+1)-x+1 = 0\)

Hint:

Separate the variables and use partial fractions on the x side.

Answer 1

The Correct Option is C

Step 1: Test the options at (1,0):
Only options that pass through $(1, 0)$ can be correct. A: $0 - 1 - 0 - 1 = -2$, no. B: $(0 - 1)(2) = -2$ and $2x = 2$, no. C: $0 + 1 + 0 - 1 = 0$, yes. D: $0 - 1 + 1 = 0$, yes.

Step 2: Test the slope:
For C, $y = \frac{1 - x}{x + 1}$, so $y' = \frac{-2}{(x+1)^2}$ and $\frac{y - 1}{x(x+1)} = \frac{-2x/(x+1)}{x(x+1)} = \frac{-2}{(x+1)^2}$. They match.
For D, $y = \frac{x - 1}{x + 1}$ has $y' = \frac{2}{(x+1)^2}$, with the opposite sign, so it fails.

Final Answer:
The curve is $xy + x + y - 1 = 0$, option (C). \[ \boxed{xy+x+y-1=0} \]