The equation of the curve passing through (1, | vedclass

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Question

(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}

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$(y - 1)(x + 1) + 2x = 0$

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(A) The slope of the curve is given by $\frac{dy}{dx} = \frac{y - 1}{x^2 + x}$.Separating the variables,we get $\frac{dy}{y - 1} = \frac{dx}{x(x + 1)}$.Using partial fractions,$\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.Integrating both sides: $\int \frac{dy}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) dx$.$\ln|y - 1| = \ln|x| - \ln|x + 1| + C$.$\ln|y - 1| = \ln|\frac{x}{x + 1}| + C$,which implies $y - 1 = k \cdot \frac{x}{x + 1}$.Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$: $0 - 1 = k \cdot \frac{1}{1 + 1} \Rightarrow -1 = \frac{k}{2} \Rightarrow k = -2$.Thus,$y - 1 = -2 \cdot \frac{x}{x + 1} \Rightarrow (y - 1)(x + 1) = -2x$.Therefore,$(y - 1)(x + 1) + 2x = 0$.

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

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