The bounded solution of a differential equation

Question:
Find the bounded solution of the following ordinary differential equation:
\[ \frac{dx}{dt} = 2x + \sin t \]
for $-\infty < t < \infty$.

Answer:
In order to obtain the general solution, consider
\[ \frac{dx}{dt}-2x_g = 0 \]
Let us put $D \equiv \frac{d}{dt}$ as an operator. Thus, the above will become:
\[ (D - 2)x_g = 0 \]
Thus, we can have
\[ x_g = Ce^{2t} \]
Now, consider the particular solution. Use the operator to express the whole equation.
\[ (D-2)x_p = \sin t \]
$\sin t$ can in general be replaced with exponential function $e^{it}$, and the imaginary part of the final result will be taken.
\begin{eqnarray*}
x_p &=& \frac{1}{D-2}e^{it}   \\
   &=&  \frac{1}{i-2}e^{it}   \\
   &=&  \frac{-1}{2-i}e^{it}  \\
   &=&  \frac{-1}{2-i}\frac{2+i}{2+i}e^{it}  \\
   &=&  \frac{-(2+i)}{4+1}e^{it}  \\
   &=&  \frac{-(2+i)}{5}e^{it}  \\
   &=&  \frac{-2-i}{5}(\cos t + i\sin t)  \\
   &=&   \left(-\frac{2}{5}\cos t + \frac{1}{5}\sin t \right) + i \left(-\frac{1}{5}\cos t - \frac{2}{5}\sin t \right)
\end{eqnarray*}
The particular solution is the imaginary part.
\[ x_p = -\frac{1}{5}\cos t - \frac{2}{5}\sin t \]
The entire solution is
\[ x = Ce^{2t}-\frac{1}{5}\cos t - \frac{2}{5}\sin t \]
The first term makes the solution diverged when $t \rightarrow \pm\infty$. Therefore, the term has to be eliminated. Namely, $C = 0$. The solution bounded is
\[ x = -\frac{1}{5}\cos t - \frac{2}{5}\sin t \]

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