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(D) The given differential equation is $\frac{dy}{dx} + y \cot x = 2 \cos x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py =

Vedclass · Accepted Answer

$2y \sin x + \cos 2x = c$

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(D) The given differential equation is $\frac{dy}{dx} + y \cot x = 2 \cos x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = 2 \cos x$. First,we find the integrating factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int \cot x dx} = e^{\ln(\sin x)} = \sin x$. The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$. Substituting the values,we get $y \sin x = \int (2 \cos x \sin x) dx + c$. Using the identity $2 \sin x \cos x = \sin 2x$,we have $y \sin x = \int \sin 2x dx + c$. Integrating $\sin 2x$,we get $y \sin x = -\frac{\cos 2x}{2} + c$. Multiplying by $2$,we obtain $2y \sin x = -\cos 2x + 2c$. Rearranging the terms,we get $2y \sin x + \cos 2x = C$ (where $C = 2c$). Thus,the correct option is $D$.

The solution of the differential equation $\frac{dy}{dx} + y \cot x = 2 \cos x$ is

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