1.The solution of the differential equation (y2+2x)dydx=y(y^2 + 2x)\frac{dy}{dx} = y(y2+2x)dxdy=y satisfies x=1x = 1x=1, y=1y = 1y=1. Then the solution isa.x=y2(1+logey)x = y^2(1 + \log_e y)x=y2(1+logey)b.y=x2(1+logex)y = x^2(1 + \log_e x)y=x2(1+logex)c.x=y2(1−logey)x = y^2(1 - \log_e y)x=y2(1−logey)d.y=x2(1−logex)y = x^2(1 - \log_e x)y=x2(1−logex)Login to continueOnly logged in users canattempt or see the solution.
