The equation of a curve passing through (1,0) | vedclass

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(B) Given the slope of the tangent is $\frac{dy}{dx} = \frac{y-1}{x^2+x}$.Separating the variables,we get $\frac{dy}{y-1} = \frac{dx}{x(x+1)}$.Using p

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

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(B) Given the slope of the tangent is $\frac{dy}{dx} = \frac{y-1}{x^2+x}$.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,$\int \frac{dy}{y-1} = \int (\frac{1}{x} - \frac{1}{x+1}) dx$.$\ln|y-1| = \ln|x| - \ln|x+1| + C$.$\ln|y-1| = \ln|\frac{x}{x+1}| + C$.Since the curve passes through $(1,0)$,substitute $x=1$ and $y=0$:$\ln|0-1| = \ln|\frac{1}{1+1}| + C \implies 0 = \ln(\frac{1}{2}) + C \implies C = \ln(2)$.Thus,$\ln|y-1| = \ln|\frac{x}{x+1}| + \ln(2) = \ln|\frac{2x}{x+1}|$.$y-1 = \frac{2x}{x+1} \implies (y-1)(x+1) = 2x$.Rearranging gives $2x - (y-1)(x+1) = 0$.

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

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