The solution of the equation frac{dy}{dx} + 2 | vedclass

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Question

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

Vedclass · Accepted Answer

$y = 2 \sin^2 x + \cos x - 2$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} + 2y 	an x = \sin x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2 	an x$ and $Q = \sin x$.The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int 2 	an x dx} = e^{2 \ln |\sec x|} = e^{\ln \sec^2 x} = \sec^2 x$.The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.$y \sec^2 x = \int \sin x \cdot \sec^2 x dx + c$.$y \sec^2 x = \int \sec x 	an x dx + c$.$y \sec^2 x = \sec x + c$.Given $y = 0$ when $x = \frac{\pi}{3}$,we substitute these values:$0 \cdot \sec^2(\frac{\pi}{3}) = \sec(\frac{\pi}{3}) + c$.$0 = 2 + c \implies c = -2$.Thus,$y \sec^2 x = \sec x - 2$.Dividing by $\sec^2 x$,we get $y = \frac{\sec x}{\sec^2 x} - \frac{2}{\sec^2 x} = \cos x - 2 \cos^2 x$.Using $\cos^2 x = 1 - \sin^2 x$,we get $y = \cos x - 2(1 - \sin^2 x) = \cos x - 2 + 2 \sin^2 x$.Therefore,$y = 2 \sin^2 x + \cos x - 2$.

The solution of the equation $\frac{dy}{dx} + 2y \tan x = \sin x$ satisfying $y = 0$ when $x = \frac{\pi}{3}$ is:

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