f(x) = E(0, inf) [((-1)^n)(x^2n)]/(2n+1)!
And I found f'(x) to be E (0,inf) [2n((-1)^n)(x^(2n-1))]/(2n+1)!
The question is:
Show that y = f(x) is a solution to the differention equation xy'+y=cosx
Not sure really where to start or how to show this. Help appreciated.
xy' + y = cosx
= x*E(0,inf) 2n(-1)^n*x^(2n-1)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!
= E(0,inf) 2n(-1)^n*x^(2n-1+1)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!
= E(0,inf) 2n(-1)^n*x^(2n)/(2n + 1)! + E(0,inf) (-1)^n*x^(2n)/(2n + 1)!}
= E(0,inf) [2n(-1)^n*x^(2n)/(2n + 1)! + (-1)^n*x^(2n)/(2n + 1)!]
= E(0,inf) [(-1)^n*x^(2n)/[(2n)!(2n + 1)]*(2n + 1)]
= E(0,inf) [(-1)^n*x^(2n)/(2n)!] = cosx