Linear First Order ODE/y' = x + y/y(0) = 1

Theorem
The linear first order ODE:
 * $(1): \quad \dfrac {\d y} {\d x} = x + y$

with initial condition:
 * $\map y 0 = 1$

has the particular solution:
 * $y = 2 e^x - x - 1$

Proof
From Linear First Order ODE: $y' = x + y$, the general solution of $(1)$ is:
 * $y = C e^x - x - 1$

Setting $y = 1$ when $x = 0$ gives:
 * $1 = C + 1$

from which $C = 2$.

Hence the result.