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(D) Given differential equation is $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$. Rearranging the terms,we get $\cos x(\frac{dy}{dx})+y=\cos^2 x$. Dividing by

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

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(D) Given differential equation is $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$. Rearranging the terms,we get $\cos x(\frac{dy}{dx})+y=\cos^2 x$. Dividing by $\cos x$,we get $\frac{dy}{dx}+y\sec x=\cos x$. This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=\sec x$ and $Q=\cos x$. The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int \sec x dx} = e^{\ln|\sec x+	an x|} = \sec x+	an x$. The general solution is $y(IF) = \int Q(IF) dx + c$. $y(\sec x+	an x) = \int \cos x(\sec x+	an x) dx + c$. $y(\sec x+	an x) = \int (1+\sin x) dx + c$. $y(\sec x+	an x) = x-\cos x+c$. Since the provided options do not match this result exactly,we re-examine the equation. If the equation was $\frac{dy}{dx} + y 	an x = \cos x$,the solution would differ. Given the structure,the correct form is $(\sec x+	an x)y = x-\cos x+c$.

The general solution of the differential equation $y+\cos x(\frac{dy}{dx})-\cos^2 x=0$ is

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