JEE Main 201815 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002Let y = y(x) be the solution of the differential equation dydx+2y=f(x),{{dy} \over {dx}} + 2y = f\left( x \right),dxdy+2y=f(x), where f(x)={1,x∈[0,1]0,otherwisef\left( x \right) = \left\{ {\matrix{ {1,} & {x \in \left[ {0,1} \right]} \cr {0,} & {otherwise} \cr } } \right.f(x)={1,0,x∈[0,1]otherwise If y(0) = 0, then y(32)y\left( {{3 \over 2}} \right)y(23) is :AOption A. e2+12e4{{{e^2} + 1} \over {2{e^4}}}2e4e2+1BOption B. 12e{1 \over {2e}}2e1COption C. e2−1e3{{{e^2} - 1} \over {{e^3}}}e3e2−1DOption D. e2−12e3{{{e^2} - 1} \over {2{e^3}}}2e3e2−1CheckSkip · show solution
