Question

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

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

Hint:

Use partial fractions for rational integrals.

Answer 1

The Correct Option is B

To find the equation of the curve that passes through the point \((1,0)\) and has a slope given by the expression \(\frac{y-1}{x^2+x}\), we need to follow these steps:

1. The slope of the curve \( \frac{dy}{dx} \) is given as \(\frac{y-1}{x^2+x}\).
2. This translates into the differential equation: \(\frac{dy}{dx} = \frac{y-1}{x^2+x}\).
3. To solve it, we can rewrite the differential equation in a separable form. Multiplying both sides by \((x^2+x)\) and integrating both sides gives us: \((x^2+x) dy = (y-1) dx\).
4. Separating variables, we have: \(\frac{dy}{y-1} = \frac{dx}{x^2+x}\).
5. The right side can be integrated using partial fractions: \(\frac{1}{x^2+x} = \frac{1}{x} - \frac{1}{x+1}\). Integrate both sides: \(\int \frac{dy}{y-1} = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx\).
6. Integrating each term, we get:
   • Left side: \(\ln{|y-1|}\)
   • Right side: \(\ln{|x|} - \ln{|x+1|} + C\) (where \( C \) is the constant of integration)
7. This integration gives us: \(\ln{|y-1|} = \ln{\left|\frac{x}{x+1}\right|} + C\).
8. Exponentiating both sides, we have: \(|y-1| = A \cdot \left|\frac{x}{x+1}\right|\) where \( A = e^C \).
9. Since the curve passes through the point \( (1, 0) \), substitute: \(0 - 1 = A \cdot \frac{1}{2}\) which gives \( A = -2 \).
10. Substituting back, we have: \(y - 1 = -2 \cdot \frac{x}{x+1}\).
11. Simplifying, the equation becomes: \((x+1)(y-1) + 2x = 0\).

Therefore, the correct equation of the curve is \((x+1)(y-1)+2x=0\), which matches the given correct answer.