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 only \(x\).

\[ \frac{dx}{dt}-2x_g = 0 \]

Let us put \(D \equiv \frac{d}{dt}\) as an operator. Thus, the above will be expressed as:

\[ (D – 2)x_g = 0 \]

Thus, we can have the general solution as

\[ x_g = Ce^{2t} \]

Now, consider the particular solution. Use the operator to express the whole equation as follows:

\[ (D-2)x_p = \sin t \]

The right hand side, \(\sin t\) can in general be replaced with exponential function \(e^{it}\), and the imaginary part of the final result will be taken as the solution.

\[  x_p = \frac{1}{D-2}e^{it}   \]

\[   =  \frac{1}{i-2}e^{it}   \]

\[   =  \frac{-1}{2-i}e^{it}  \]

\[   =  \frac{-1}{2-i}\frac{2+i}{2+i}e^{it}  \]

\[   =  \frac{-(2+i)}{4+1}e^{it}  \]

\[   =  \frac{-(2+i)}{5}e^{it}  \]

\[   =  \frac{-2-i}{5}(\cos t + i\sin t)  \]

\[   =   \left(-\frac{2}{5}\cos t + \frac{1}{5}\sin t \right) + i \left(-\frac{1}{5}\cos t – \frac{2}{5}\sin t \right)  \]

The particular solution is that of the imaginary part.

\[ x_p = -\frac{1}{5}\cos t – \frac{2}{5}\sin t \]

The entire solution will become

\[ x = Ce^{2t}-\frac{1}{5}\cos t – \frac{2}{5}\sin t \]

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

\[ x = -\frac{1}{5}\cos t – \frac{2}{5}\sin t \]