The solution of the differential equation y' | vedclass

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(D) The given differential equation is $y' - y 	an x = -2 \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where

Vedclass · Accepted Answer

None of these

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(D) The given differential equation is $y' - y 	an x = -2 \sin x$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -	an x$ and $Q = -2 \sin x$. The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P \, dx} = e^{-\int 	an x \, dx} = e^{-\ln |\sec x|} = e^{\ln |\cos x|} = \cos x$. The solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) \, dx + c$. Substituting the values,we get $y \cos x = \int (-2 \sin x \cos x) \, dx + c$. $y \cos x = -\int \sin 2x \, dx + c$. $y \cos x = \frac{\cos 2x}{2} + c$. Multiplying by $\sec x$,we get $y = \frac{\cos 2x}{2 \cos x} + c \sec x$. Since none of the given options match this result,the correct option is $D$.

The solution of the differential equation $y' = y \tan x - 2 \sin x$ is

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