Let y=y(x) be the solution of the differential equation x (x) dy dx + y = 2 (x) for x > 1 . If y(e) = 5 , then the minimum value of y(x) for x > 1 is equal to:

Step-by-step solution

The given differential equation is x (x) dy dx + y = 2 (x) . Dividing by x (x) , we get: dy dx + 1 x (x) y = 2 x This is a linear differential equation of the form dy dx + P(x)y = Q(x) , where P(x) = 1 x (x) . The integrating factor (I.F.) is: I.F. = e^ 1 x (x) dx Let (x) = t , then 1 x dx = dt . 1 t dt = (t) = ( (x)) So, I.F. = e^ ( (x)) = (x) . Multiplying the equation by the I.F. and integrating, we get: y (x) = 2 x (x) dx Using the same substitution t = (x) , the integral becomes: 2t dt = t^2 + C = ( (x))^2 + C