A curve y = f(x) in the first quadrant has the property that the slope of the tangent at any point (x, y) on it is given by y x + 2x x . If the curve passes through the point ( 2 , 0 ) , then the value of f ( 6 ) is :

Step-by-step solution

The given condition can be written as the differential equation: dy dx = y x + 2x x dy dx - y x = 2x x This is a linear differential equation of the form dy dx + P(x)y = Q(x) , where P(x) = - x and Q(x) = 2x x . The integrating factor (I.F.) is: I.F. = e^ - x dx = e^ - | x| = x Multiplying the differential equation by the integrating factor, we get: d dx (y x) = 2x x x = 2x Integrating both sides with respect to x : y x = x^2 + C The curve passes through ( 2 , 0 ) , so substitute x = 2 and y = 0 : 0 ( 2 ) = ( 2 )^2