Let y = y(x) be the solution of the differential equation (x^2+1)^2 dy dx + 2x(x^2+1)y = x , subject to the initial condition y(0) = 1 . Then the value of _ x 0 2y(x) - 2 x^2 is equal to :

Step-by-step solution

The given differential equation is: (x^2+1)^2 dy dx + 2x(x^2+1)y = x Dividing the entire equation by x^2+1 , we get: (x^2+1) dy dx + 2xy = x x^2+1 Notice that the left hand side is the exact derivative of y(x^2+1) : d dx (y(x^2+1)) = x x^2+1 Integrating both sides with respect to x : y(x^2+1) = 1 2 (x^2+1) + C Using the initial condition y(0) = 1 : 1(0^2+1) = 1 2 (0+1) + C C = 1 So, the solution is: y(x^2+1) = 1 2 (x^2+1) + 1 y(x) = (x^2+1) + 2 2(x^2+1) Now, we need to evaluate the limit: L = _ x 0 2y(x) - 2 x^2 Su