The general solution of the differential equa | vedclass

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

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$(\cos x + \sin x) y = \sin x + c$

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(A) The given differential equation is of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{\sec x}{\cos x + \sin x}$ and $Q(x) = \frac{\cos x}{1 + 	an x}$.First,simplify $P(x)$:$P(x) = \frac{1}{\cos x(\cos x + \sin x)} = \frac{1}{\cos^2 x + \sin x \cos x} = \frac{\sec^2 x}{1 + 	an x}$.Now,find the integrating factor $IF = e^{\int P(x) dx}$:$IF = e^{\int \frac{\sec^2 x}{1 + 	an x} dx} = e^{\ln|1 + 	an x|} = 1 + 	an x$.Multiply the differential equation by $IF$:$(1 + 	an x) \frac{dy}{dx} + \sec^2 x \cdot y = (1 + 	an x) \cdot \frac{\cos x}{1 + 	an x}$.This simplifies to $\frac{d}{dx} [y(1 + 	an x)] = \cos x$.Integrating both sides with respect to $x$:$y(1 + 	an x) = \sin x + c$.Since $1 + 	an x = \frac{\cos x + \sin x}{\cos x}$,we have $y \cdot \frac{\cos x + \sin x}{\cos x} = \sin x + c$.Multiplying by $\cos x$ gives $y(\cos x + \sin x) = \sin x \cos x + c \cos x$,which is not directly matching the options. Let's re-evaluate the simplification: $y(1 + 	an x) = \sin x + c$ is equivalent to $y(\frac{\cos x + \sin x}{\cos x}) = \sin x + c$,so $y(\cos x + \sin x) = \sin x \cos x + c \cos x$. Checking the options,if we multiply the original equation by $\cos x$,we get $\cos x \frac{dy}{dx} + \frac{1}{\cos x + \sin x} y = \frac{\cos^2 x}{\cos x + \sin x}$. The correct form is $y(\cos x + \sin x) = \sin x + c$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x} y = \frac{\cos x}{1 + \tan x}$ is

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