1.The solution of (1+x2)dy+2xy dx−4x2 dx=0(1 + x^2)dy + 2xy\,dx - 4x^2\,dx = 0(1+x2)dy+2xydx−4x2dx=0 isa.3x(1+y2)=4y3+c3x(1 + y^2) = 4y^3 + c3x(1+y2)=4y3+cb.3y(1+x2)=4x3+c3y(1 + x^2) = 4x^3 + c3y(1+x2)=4x3+cc.3x(1−y2)=4y3+c3x(1 - y^2) = 4y^3 + c3x(1−y2)=4y3+cd.3y(1+y2)=4x3+c3y(1 + y^2) = 4x^3 + c3y(1+y2)=4x3+cLogin to continueOnly logged in users canattempt or see the solution.
