If $ f(x) $ satisfies the relation $ f(x) = e^x + \int_0^x (y + xe^x) f(y),dy $, then $ e + f(0) $ is equal to ______

$ f(x) = e^x + \int_0^x y f(y),dy + xe^x \int_0^x f(y),dy $ $ f(x) = e^x + A + Bxe^x $ $ A = \int_0^x y f(y),dy = \int_0^x y(A + e^y + Bye^y),dy $ $ A = \frac{A}{2} + 0(-1) + B(e - 1) $ $ \frac{A}{2} + B(1 - e) = 1 $ $ B = \int_0^1 f(y),dy $ $ B = \int_0^1 (e^y + A + Bye^y),dy $ $ B = (e - 1) + A + B(0 - 1) $ $ B = e - 1 + A + B(0 - (-1)) $ $ B = e - 1 + A + B \Rightarrow A = 1 - e $ $ f(x) = e^x + A + Bxe^x $ $ f(0) = 1 + A + B\cdot 0 = 1 + A = 2 - e $ $ e + f(0) = 2 $