MHT CET · Maths · Differential Equations
The equation of a curve passing through \((1,0)\) and having slope of tangent at any point \((\mathrm{x}, \mathrm{y})\) of the curve as \(\frac{\mathrm{y}-1}{x^2+x}\) is
- A \(2(\mathrm{y}-1)+x(x+1)=0\)
- B \(2 x-(y-1)(x+1)=0\)
- C \(2 x+(x+1)(y-1)=0\)
- D \(2 x(y-1)+(x+1)=0\)
Answer & Solution
Correct Answer
(C) \(2 x+(x+1)(y-1)=0\)
Step-by-step Solution
Detailed explanation
\(\frac{dy}{dx} = \frac{y-1}{x(x+1)}\) \(\int \frac{dy}{y-1} = \int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx\)
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