Let a curve C be the solution of the differential equation 2x dy dx - y = 0 . If C passes through the point (2, 2 3 ) , then the area of the region in the first quadrant bounded by C , the x -axis and the circle x^2 + y^2 = 16 is equal to

Step-by-step solution

First, solve the given differential equation: 2x dy dx = y dy y = dx 2x Integrating both sides: y = 1 2 x + C y = k x y^2 = k^2 x Let k^2 = c , so the curve is a parabola y^2 = cx . Since it passes through (2, 2 3 ) : (2 3 )^2 = c(2) 12 = 2c c = 6 The equation of the curve C is y^2 = 6x . In the first quadrant, y = 6x . Next, find the point of intersection of the parabola y^2 = 6x and the circle x^2 + y^2 = 16 in the first quadrant: x^2 + 6x - 16 = 0 (x + 8)(x - 2) = 0 Since x > 0 in the first quadrant, x = 2 . At