Let y = f(x) be a solution to the differential equation x x dy dx + y = 2x^2 x defined for x > 1 . If _ x 1^+ f(x) exists and is finite, then the value of f( e ) is equal to

Step-by-step solution

Given differential equation is x x dy dx + y = 2x^2 x Dividing by x x , we get dy dx + 1 x x y = 2x This is a linear differential equation of the form dy dx + P(x)y = Q(x) . Integrating Factor (I.F.) = e^ 1 x x dx = e^ ( x) = x The general solution is given by: y x = 2x x dx Using integration by parts: y x = x 2x dx - ( 1 x x^2 ) dx y x = x^2 x - x^2 2 + C f(x) = x^2 + C - x^2 2 x For _ x 1^+ f(x) to be finite, the numerator must approach 0 as x 1^+ . C - 1 2 = 0 C = 1 2 Thus, f(x) = x^2 + 1 - x^2 2 x Substituting