Solutions of $y'-2y=1$ I claim the solution of $y'-2y=1$, is $$\phi(x)=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}=e^{2x}\left(-\frac{1}{2}e^{-2x}\right)+ce^{2x}=-\frac{1}{2}+ce^{2x}$$
where $x_0$ is a fixed point on the interval where the function $b(x)=1$ is continuous.
Apparently the correct solution is $-\frac{1}{2}+ce^{-2x}$.
A second question is, why can I evaluate the integral above and ignore the term which involves $\frac{1}{2}e^{-2x_0}$ after applying FTC? Somehow it vanishes, and this integral is basically just the primitive evaluated at $x$.
 A: We can check whether your solution works:
\begin{align}
\phi(x)&=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}\\
\phi'(x) &=2e^{2x}\int_{x_0}^xe^{-2t}dt + e^{2x}e^{-2x} + 2ce^{2x}=2e^{2x}\int_{x_0}^xe^{-2t}dt+1+2ce^{2x}.
\end{align}
Now,
\begin{align}
\phi'(x) - 2\phi(x) &= 2ce^{2x}=2e^{2x}\int_{x_0}^xe^{-2t}dt+1+2ce^{2x} - 2e^{2x}\int_{x_0}^xe^{-2t}dt +-2ce^{2x} = 1\\
\end{align}
which means that your solution is correct.  Now, to simplify $\phi(x)$:
\begin{align}
\phi(x)&=e^{2x}\int_{x_0}^xe^{-2t}dt +ce^{2x}\\
&=-\frac{e^{2x}}{2}\left(e^{-2t}\right)\Bigg|_{x_{0}}^{x} + ce^{2x}\\
&=-\frac{e^{2x}}{2}\left(e^{-2x} - e^{-2x_{0}}\right) + ce^{2x}\\
&=-\frac{1}{2}+\frac{e^{-2x_{0}}}{2}e^{2x} + ce^{2x}\\
&=-\frac{1}{2} + \left(\frac{e^{-2x_{0}}}{2} + c\right)e^{2x}\\
&=-\frac{1}{2} + Ce^{2x},
\end{align}
where $C = \frac{e^{-2x_{0}}}{2} + c$ is a constant.  So it seems like there is a typo in the answer you were given, and the reason that you can seemingly ignore the term with $\frac{1}{2}e^{-2x_0}$ is because it gets absorbed into the integration constant.
