Question

The equation of a curve passing through (1,0) and having slope of tangent at any point (x, y) of the curve as $\frac{y-1}{x^2+x}$ is ______.

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

Hint:

Using $\ln|c|$ as the integration constant instead of just $+C$ makes solving logarithmic differential equations much cleaner, allowing you to merge all terms into a single fraction instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The slope of the tangent is $\frac{dy}{dx}$. We need to solve the differential equation $\frac{dy}{dx} = \frac{y-1}{x^2+x}$ using variable separable method.

Step 2: Formula Application:
$\frac{dy}{y-1} = \frac{dx}{x(x+1)}$

Step 3: Explanation:
Integrate both sides: $\int \frac{dy}{y-1} = \int \left( \frac{1}{x} - \frac{1}{x+1} \right) dx$. $\log(y-1) = \log x - \log(x+1) + \log c \implies y-1 = \frac{cx}{x+1}$. The curve passes through $(1, 0)$: $0 - 1 = \frac{c(1)}{1+1} \implies -1 = \frac{c}{2} \implies c = -2$. Equation: $y-1 = \frac{-2x}{x+1} \implies (y-1)(x+1) = -2x \implies 2x + (y-1)(x+1) = 0$. (Note: Re-checking signs for option alignment). If $y-1$ is treated as $(1-y)$ for log, we get option (b).

Step 4: Final Answer:
The equation is $2x + (y-1)(x+1) = 0$.