Let y = y(x) be the solution of the differential equation dy dx + y x = x x 1 + ^2 x for x [0, 2 ) , satisfying y(0) = - 4 . Then the value of the integral ₀^ 2 y(x) , dx is equal to

Step-by-step solution

The given differential equation is a linear differential equation of the form dy dx + P(x)y = Q(x) , with P(x) = x . The integrating factor (I.F.) is e^ x , dx = e^ ( x) = x . The general solution is given by: y x = x x 1 + ^2 x x , dx + C y x = x 1 + ^2 x , dx + C Let x = u , then - x , dx = du . -du 1 + u^2 = - (u) = - ( x) . Thus, y x = - ( x) + C . Using the initial condition y(0) = - 4 : (- 4 ) (0) = - ( 0) + C - 4 = - (1) + C - 4 = - 4 + C C = 0 . So, y(x) = - x ( x) . We need to evaluate the integral I = ₀^