A curve y = f(x) passes through the origin (0, 0) . The slope of the tangent to the curve at any point (x,y) is given by 2 (x^2+1)^2 - 2xy x^2+1 . Then the value of _ x x^2 f(x) is

Step-by-step solution

The slope of the tangent to the curve is given by dy dx . According to the problem: dy dx = 2 (x^2+1)^2 - 2xy x^2+1 dy dx + ( 2x x^2+1 )y = 2 (x^2+1)^2 This is a linear differential equation with integrating factor (I.F.): I.F. = e^ 2x x^2+1 dx = e^ (x^2+1) = x^2+1 . Multiplying by the I.F., the solution is given by: y(x^2+1) = 2 (x^2+1)^2 (x^2+1) dx y(x^2+1) = 2 x^2+1 dx y(x^2+1) = 2 ⁻¹(x) + C Since the curve passes through the origin (0,0) : 0(0^2+1) = 2 ⁻¹(0) + C C = 0 . Therefore, the equation of the curve is: