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(D) Given the differential equation: $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$.Let $v = y - x$. Then,differentiating with respect to $x$,we get $\frac

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$x + \sec(y-x) = c$

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(D) Given the differential equation: $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$.Let $v = y - x$. Then,differentiating with respect to $x$,we get $\frac{dv}{dx} = \frac{dy}{dx} - 1$,which implies $\frac{dy}{dx} = 1 + \frac{dv}{dx}$.Substituting these into the original equation:$1 + \frac{dv}{dx} = 1 - \cos v \cot v$$\frac{dv}{dx} = -\cos v \left( \frac{\cos v}{\sin v} \right) = -\frac{\cos^2 v}{\sin v}$.Separating the variables:$-\int \frac{\sin v}{\cos^2 v} dv = \int dx$.Using the substitution $u = \cos v$,$du = -\sin v dv$,the integral becomes:$\int \frac{du}{u^2} = x + c$$-\frac{1}{u} = x + c$$-\frac{1}{\cos v} = x + c$$-\sec v = x + c$$x + \sec(y-x) = c$.

The solution of the differential equation $\frac{dy}{dx} = 1 - \cos(y-x) \cot(y-x)$ is

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