Let y=Y(x) be the solution of the differential equation d y d x +y x=2 x+x^2 x , x ( - 2 , 2 ) , such that Y(0)=1 , then

Step-by-step solution

d y d x +y x=2 x+x^2 x, x ( - 2 , 2 ) On comparing with form d y d x +P y=Q , where P and Q are the functions of x . P= x and Q=2 x+x^2 x Hence, I F=e^ x d x =e^ x = x y I F= Q I F+C aligned & y x= (2 x+x^2 x ) x+C & y x=2 x x d x+ x^2 x ^2 x d x+C & y x= 2 x x d x+ x^2 x x d x+C & y x=x^2 x+C & y=x^2+C x...(i) aligned Now, when x = 0, y = 1 or y(0) = 1 From Eq. (i), we get 1=0+C 0^ aligned & 1=C & y=x^2+ x & y^ =2 x- x aligned aligned & y^ ( 4 )=2 4 - 4 = 2 - 1 2 & y^ ( - 4 )=2 ( - 4 )- ( - 4 )= - 2 + 1 2 & aligne