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 Question:
The slope of a tangent is the first derivative $\frac{dy}{dx}$. We are given a differential equation and an initial condition point $(1,0)$. We must solve it to find the explicit curve equation.

Step 2: Detailed Explanation:
The differential equation is given by:
$\frac{dy}{dx} = \frac{y-1}{x^2+x}$
Separate the variables ($y$ terms on the left, $x$ terms on the right):
$\frac{dy}{y-1} = \frac{dx}{x^2+x}$
Factor the denominator on the right side:
$\frac{dy}{y-1} = \frac{dx}{x(x+1)}$
Use partial fractions to split the right side:
$\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x+1}$
So the equation becomes:
$\frac{dy}{y-1} = \left( \frac{1}{x} - \frac{1}{x+1} \right) dx$
Integrate both sides:
$\int \frac{1}{y-1} dy = \int \left( \frac{1}{x} - \frac{1}{x+1} \right) dx$
$\ln|y-1| = \ln|x| - \ln|x+1| + \ln|c|$
Combine the logarithms on the right side using $\ln A - \ln B = \ln(A/B)$ and $\ln A + \ln B = \ln(AB)$:
$\ln|y-1| = \ln \left| \frac{cx}{x+1} \right|$
Remove the natural logarithm from both sides:
$y-1 = \frac{cx}{x+1}$ --- (Equation of the family of curves)
Now apply the initial condition: the curve passes through the point $(1, 0)$.
Substitute $x=1, y=0$:
$0 - 1 = \frac{c(1)}{1+1}$
$-1 = \frac{c}{2} \implies c = -2$
Substitute $c = -2$ back into the curve equation:
$y - 1 = \frac{-2x}{x+1}$
Multiply both sides by $(x+1)$ to remove the fraction:
$(y - 1)(x + 1) = -2x$
Rearrange to match the options (bring $-2x$ to the left side):
$2x + (x + 1)(y - 1) = 0$

Step 3: Final Answer:
The equation is $2x + (x+1)(y-1) = 0$, matching option (c).