The equation of the curve whose slope is frac | vedclass

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(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x^2 + x} = \frac{y - 1}{x(x + 1)}$.Separating variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x

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

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(B) Given the slope $\frac{dy}{dx} = \frac{y - 1}{x^2 + x} = \frac{y - 1}{x(x + 1)}$.Separating variables: $\int \frac{dy}{y - 1} = \int \frac{dx}{x(x + 1)}$.Using partial fractions: $\frac{1}{x(x + 1)} = \frac{1}{x} - \frac{1}{x + 1}$.Integrating both sides: $\ln|y - 1| = \ln|x| - \ln|x + 1| + C = \ln|\frac{x}{x + 1}| + C$.Since the curve passes through $(1, 0)$, substitute $x = 1, y = 0$: $\ln|0 - 1| = \ln|\frac{1}{1 + 1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.Thus, $\ln|y - 1| = \ln|\frac{x}{x + 1}| + \ln(2) = \ln|\frac{2x}{x + 1}|$.Taking exponents: $y - 1 = \frac{2x}{x + 1} \implies (y - 1)(x + 1) = 2x$.

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

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