The general solution of the differential equa | vedclass

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(D) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \sec x \operatorname{cosec} x = \frac{1}{\cos x \sin x

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$2y 	an x = \sin^2 x + c$

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(D) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \sec x \operatorname{cosec} x = \frac{1}{\cos x \sin x} = \frac{2}{\sin 2x}$ and $Q(x) = \cos^2 x$. First,we find the integrating factor $(IF)$: $IF = e^{\int P(x) dx} = e^{\int \frac{1}{\sin x \cos x} dx} = e^{\int \frac{\sec^2 x}{	an x} dx} = e^{\ln|	an x|} = 	an x$. The general solution is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + c$. $y 	an x = \int \cos^2 x \cdot 	an x dx + c$. $y 	an x = \int \cos^2 x \cdot \frac{\sin x}{\cos x} dx + c = \int \sin x \cos x dx + c$. $y 	an x = \frac{1}{2} \int \sin 2x dx + c = -\frac{1}{4} \cos 2x + c$. Alternatively,using $\int \sin x \cos x dx = \frac{\sin^2 x}{2} + c$,we get $y 	an x = \frac{\sin^2 x}{2} + c$,which implies $2y 	an x = \sin^2 x + c$.

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

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