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(D) The given differential equation is:$\frac{dy}{dx} + \csc^2 (x) \cdot y = \csc^2 (x) \cdot \cot x . . . . . . . . . . (1)$Comparing this with the l

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$y e^{- \cot x} = (1 + \cot x) e^{- \cot x} + c$

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(D) The given differential equation is:$\frac{dy}{dx} + \csc^2 (x) \cdot y = \csc^2 (x) \cdot \cot x . . . . . . . . . . (1)$Comparing this with the linear differential equation form $\frac{dy}{dx} + Py = Q$,we get:$P = \csc^2 x$$Q = \csc^2 x \cdot \cot x$Now,find the Integrating Factor $(IF)$:$IF = e^{\int P \ dx} = e^{\int \csc^2 x \ dx} = e^{- \cot x}$The general solution is given by:$y \cdot IF = \int Q \cdot IF \ dx + C$$y \cdot e^{- \cot x} = \int \csc^2 x \cdot \cot x \cdot e^{- \cot x} \ dx + C$Let $t = \cot x$,then $dt = - \csc^2 x \ dx$,which implies $\csc^2 x \ dx = - dt$.Substituting these into the integral:$y \cdot e^{- \cot x} = \int t \cdot e^{- t} (- dt) = - \int t e^{- t} \ dt$Using integration by parts $\int u \ dv = uv - \int v \ du$ where $u = t$ and $dv = e^{- t} \ dt$:$- \int t e^{- t} \ dt = - [t (- e^{- t}) - \int (- e^{- t}) \ dt] = - [- t e^{- t} - e^{- t}] + C = t e^{- t} + e^{- t} + C$Substituting $t = \cot x$ back:$y \cdot e^{- \cot x} = e^{- \cot x} (\cot x + 1) + C$.

Find the solution of the differential equation given below:
$\frac{dy}{dx} + y \cdot \csc^2 (x) = \csc^2 (x) \cdot \cot (x)$

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Find the solution of the differential equation given below:$\frac{dy}{dx} + y \cdot \csc^2 (x) = \csc^2 (x) \cdot \cot (x)$