MathematicsEasy38×since 2002Q3506The solution of the differential equation dydx=x+yx{{dy} \over {dx}} = {{x + y} \over x}dxdy=xx+y satisfying the condition y(1)=1y(1)=1y(1)=1 is :Ay=lnx+xy = \ln x + xy=lnx+xBy=xlnx+x2y = x\ln x + {x^2}y=xlnx+x2Cy=xe(x−1) y = x{e^{\left( {x - 1} \right)}}\,y=xe(x−1)Dy=x lnx+xy = x\,\ln x + xy=xlnx+xCheck answerSkip