If dydx+y.tanx=sin2x and y(0)=1then y(π)=____

dydx+ytanx=sin2x I.F.=e∫tanx dx=secxSolution : ysecx=∫(sin2x)(secx)dxy(secx)=2(−cosx)+c →1y = 1, where x = 0 1(1)=−2+c⇒c=3eq.1⇒ysecx=−2cosx+3 →2put x=π in eq2, we sety(−1)=−2(−1)+3 ∴y(π)=−5

If dydx+y.tanx=sin2x and y(0)=1then y(π)=____

dydx+ytanx=sin2x I.F.=e∫tanx dx=secxSolution : ysecx=∫(sin2x)(secx)dxy(secx)=2(−cosx)+c →1y = 1, where x = 0 1(1)=−2+c⇒c=3eq.1⇒ysecx=−2cosx+3 →2put x=π in eq2, we sety(−1)=−2(−1)+3 ∴y(π)=−5

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If $\frac{d y}{d x} + y . \tan x = \sin 2 x$ and $y (0) = 1$ then $y (π) =$ ____

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-1

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Detailed Solution

$\frac{d y}{d x} + y \tan x = \sin 2 x I . F . = e^{\int \tan x d x} = \sec x$
Solution : $y \sec x = \int (\sin 2 x) (\sec x) d x$
$y (\sec x) = 2 (- \cos x) + c \to 1$
y = 1, where x = 0
$1 (1) = - 2 + c \Rightarrow c = 3$
$e q .1 \Rightarrow y \sec x = - 2 \cos x + 3 \to 2$
put $x = π$ in eq2, we set
$y (- 1) = - 2 (- 1) + 3 ∴ y (π) = - 5$