JEE Main 20199 Apr, Morning shiftMathematicsMediumDifferential Equations163× since 2002The solution of the differential equation xdydx+2yx{{dy} \over {dx}} + 2yxdxdy+2y = x² (x ≠\ne= 0) with y(1) = 1, is :AOption A. y=45x3+15x2y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}y=54x3+5x21BOption B. y=34x2+14x2y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}y=43x2+4x21COption C. y=x24+34x2y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}y=4x2+4x23DOption D. y=x35+15x2y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}y=5x3+5x21CheckSkip · show solution
