The equation of the curve passing through (3, | vedclass

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Question

(C) Given the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int (x + x^{-2}) dx$$y

Vedclass · Accepted Answer

$6xy = 3x^3 + 29x - 6$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int (x + x^{-2}) dx$$y = \frac{x^2}{2} - \frac{1}{x} + c$Since the curve passes through $(3, 9)$,we substitute $x = 3$ and $y = 9$ to find $c$:$9 = \frac{3^2}{2} - \frac{1}{3} + c$$9 = \frac{9}{2} - \frac{1}{3} + c$$9 = \frac{27 - 2}{6} + c$$9 = \frac{25}{6} + c$$c = 9 - \frac{25}{6} = \frac{54 - 25}{6} = \frac{29}{6}$Substituting $c$ back into the equation:$y = \frac{x^2}{2} - \frac{1}{x} + \frac{29}{6}$Multiply the entire equation by $6x$:$6xy = 3x^3 - 6 + 29x$$6xy = 3x^3 + 29x - 6$.

The equation of the curve passing through $(3, 9)$ which satisfies the differential equation $\frac{dy}{dx} = x + \frac{1}{x^2}$ is

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