Let y = y(x) be the solution of the differential equation dy dx - y x = 2x for x (0, ) . If the minimum value of y(x) on this interval is - 1 8 , then y ( 2 ) is equal to

Step-by-step solution

The given differential equation is dy dx - y x = 2x . This is a linear differential equation of the form dy dx + P(x)y = Q(x) , where P(x) = - x and Q(x) = 2x . The integrating factor (I.F.) is e^ - x , dx = e^ - ( x) = 1 x = x . Multiplying the differential equation by the I.F. and integrating, we get the general solution: y x = ( 2x)( x) , dx + C y x = 2 x x x , dx + C y x = 2 x , dx + C y x = 2 x + C y(x) = 2 ^2 x + C x Let t = x . Since x (0, ) , we have t (0, 1] . The function becomes a quadratic in t : f(t) =