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: Understanding the Concept:
The slope of the curve at any point is \(\frac{dy}{dx}\). Setting \(\frac{dy}{dx} = \frac{y - 1}{x^2 + x}\) gives a differential equation with separable variables.

Step 2: Key Formula or Approach:
\(\frac{dy}{y - 1} = \frac{dx}{x(x + 1)} = \left(\frac1x - \frac1{x + 1}\right)dx\).

Step 3: Detailed Explanation:
Integrate: \(\log|y - 1| = \log|x| - \log|x + 1| + \log|K|\), so \(y - 1 = \frac{Kx}{x + 1}\).
The curve passes through \((1, 0)\): \(-1 = \frac K2\), so \(K = -2\).
\((y - 1)(x + 1) = -2x\), i.e. \(xy + y - x - 1 + 2x = 0\).
\[ xy + x + y - 1 = 0 \]

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