If the gradient of the tangent at any point ( | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin^2\left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Let

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$y = x \cot^{-1}(\log_e ex)$

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(C) Given the differential equation $\frac{dy}{dx} = \frac{y}{x} - \sin^2\left( \frac{y}{x} \right)$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $v + x \frac{dv}{dx} = v - \sin^2 v$.This simplifies to $x \frac{dv}{dx} = -\sin^2 v$,or $-\csc^2 v \, dv = \frac{dx}{x}$.Integrating both sides: $-\int \csc^2 v \, dv = \int \frac{dx}{x}$.This gives $\cot v = \log_e x + C$,so $\cot\left( \frac{y}{x} \right) = \log_e x + C$.The curve passes through $\left( 1, \frac{\pi}{4} \right)$,so $\cot\left( \frac{\pi/4}{1} \right) = \log_e 1 + C$.Since $\cot(\pi/4) = 1$ and $\log_e 1 = 0$,we get $C = 1$.Thus,$\cot\left( \frac{y}{x} \right) = \log_e x + 1 = \log_e x + \log_e e = \log_e(ex)$.Therefore,$\frac{y}{x} = \cot^{-1}(\log_e ex)$,which implies $y = x \cot^{-1}(\log_e ex)$.

If the gradient of the tangent at any point $(x, y)$ of a curve which passes through the point $\left( 1, \frac{\pi}{4} \right)$ is $\left\{ \frac{y}{x} - \sin^2\left( \frac{y}{x} \right) \right\}$,then the equation of the curve is:

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