JEE Main 202127 Aug, Morning shiftMathematicsEasyDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation dydx=2(y+2sinx−5)x−2cosx{{dy} \over {dx}} = 2(y + 2\sin x - 5)x - 2\cos xdxdy=2(y+2sinx−5)x−2cosx such that y(0) = 7. Then y(π\piπ) is equal to :AOption A. 2eπ2+52{e^{{\pi ^2}}} + 52eπ2+5BOption B. eπ2+5{e^{{\pi ^2}}} + 5eπ2+5COption C. 3eπ2+53{e^{{\pi ^2}}} + 53eπ2+5DOption D. 7eπ2+57{e^{{\pi ^2}}} + 57eπ2+5CheckSkip · show solution
