The equation of a curve passing through left( | vedclass

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

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$xy = x^2 + x + 1$

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(B) Given the gradient of the curve is $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$,we get:$y = \int \left( 1 - \frac{1}{x^2} \right) dx = x + \frac{1}{x} + c$.Since the curve passes through the point $\left( 2, \frac{7}{2} \right)$,we substitute these values into the equation:$\frac{7}{2} = 2 + \frac{1}{2} + c$.$\frac{7}{2} = \frac{5}{2} + c \implies c = 1$.Thus,the equation of the curve is $y = x + \frac{1}{x} + 1$.Multiplying by $x$,we get $xy = x^2 + 1 + x$,which is $xy = x^2 + x + 1$.

The equation of a curve passing through $\left( 2, \frac{7}{2} \right)$ and having gradient $1 - \frac{1}{x^2}$ at $(x, y)$ is:

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