If frac{dy}{dx} + y tan x = sin 2x and y(0) = | vedclass

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

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$-5$

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(C) The given differential equation is $\frac{dy}{dx} + y 	an x = \sin 2x$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 	an x$ and $Q = \sin 2x$.The integrating factor $(IF)$ is $e^{\int P dx} = e^{\int 	an x dx} = e^{\ln(\sec x)} = \sec x$.The general solution is $y(IF) = \int Q(IF) dx + c$.$y \sec x = \int \sin 2x \sec x dx + c$.$y \sec x = \int (2 \sin x \cos x) \sec x dx + c$.$y \sec x = 2 \int \sin x dx + c$.$y \sec x = -2 \cos x + c$  .....$(1)$.Given $y(0) = 1$,substitute $x = 0$ and $y = 1$ into $(1)$:$1 \cdot \sec(0) = -2 \cos(0) + c \Rightarrow 1(1) = -2(1) + c \Rightarrow c = 3$.Substituting $c = 3$ into $(1)$,we get $y \sec x = -2 \cos x + 3$.To find $y(\pi)$,substitute $x = \pi$:$y \sec(\pi) = -2 \cos(\pi) + 3$.$y(-1) = -2(-1) + 3$.$-y = 2 + 3 = 5$.$y = -5$.

If $\frac{dy}{dx} + y \tan x = \sin 2x$ and $y(0) = 1$,then $y(\pi)$ is equal to

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