Let the solution curve y = f(x) of the differential equation x x dy dx + y = k x , x > 1 pass through the point (e, 1) . If the tangent to the curve at x = e is parallel to the x-axis, then the value of f(e^2) is equal to

Step-by-step solution

Given differential equation is x x dy dx + y = k x Dividing by x x , we get dy dx + 1 x x y = k x 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 = k x x , dx + C y x = k 2 ( x)^2 + C Since the curve passes through (e, 1) , we substitute x = e, y = 1 : 1 e = k 2 ( e)^2 + C 1 = k 2 + C We are given that the tangent at x = e is parallel to the x-axis, so dy dx = 0 at x = e . Substituting x = e, y = 1,