A curve passes through the point (0, ) . The slope of the tangent to the curve at any point (x, y) on it is given by y x + 2y^3 y . The x -coordinate of the point on the curve whose y -coordinate is 2 is equal to :

Step-by-step solution

The slope of the tangent is given by dy dx = y x + 2y^3 y . Taking the reciprocal, we get a linear differential equation in x : dx dy = x + 2y^3 y y dx dy - 1 y x = 2y^2 y The integrating factor (I.F.) is e^ - 1 y dy = e^ - y = 1 y . Multiplying the differential equation by the I.F., we get : 1 y dx dy - 1 y^2 x = 2y y d dy ( x y ) = 2y y Integrating both sides with respect to y : x y = 2y y , dy Using integration by parts : x y = 2 ( y(- y) - 1 (- y) , dy ) x y = -2y y + 2 y + C The curve passes through (0, ) , so