The equation of the curve passing through lef | vedclass

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(B) Given the slope of the curve is $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int \left(1 - \frac{1}{

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$xy = x^2 + 2x + 1$

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(B) Given the slope of the curve is $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int \left(1 - \frac{1}{x^2}\right) dx$$y = x + \frac{1}{x} + C$Since the curve passes through $\left(2, \frac{9}{2}\right)$,substitute $x = 2$ and $y = \frac{9}{2}$:$\frac{9}{2} = 2 + \frac{1}{2} + C$$\frac{9}{2} = \frac{5}{2} + C$$C = \frac{9}{2} - \frac{5}{2} = 2$Thus,the equation is $y = x + \frac{1}{x} + 2$.Multiplying by $x$,we get $xy = x^2 + 1 + 2x$,which simplifies to $xy = x^2 + 2x + 1$.

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

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