A curve y = f(x) passes through the point (0, 2) . If the slope of the tangent to the curve at any point (x, y) in the interval (- 2 , 2 ) is given by x x - y x , then the value of f ( 3 ) is equal to

Step-by-step solution

Given that the slope of the tangent is dy dx = x x - y x . Rearranging this, we get a linear differential equation: dy dx + y x = x x The integrating factor (I.F.) is: I.F. = e^ x , dx = e^ _e( x) = x Multiplying the differential equation by the I.F. and integrating, we get: y x = x ^2 x , dx Using integration by parts on the right side: y x = x x - 1 x , dx y x = x x - _e| x| + C It is given that the curve passes through (0, 2) , so f(0) = 2 . Substituting x = 0 and y = 2 : 2 (0) = 0 - _e| (0)| + C 2(1) = 0 - 0 +