Answer
$y = \frac{9}{10}e^x\cos{2x} -\frac{1}{20}e^x\sin{2x} + \frac{1}{5}\sin{x} + \frac{1}{10}\cos{x}$
Work Step by Step
$y'' - 2y' + 5y = \sin{x}$
First, solve the homogenous equation. This yields characteristic equation:
$r^2 - 2r +5 = 0$
$ \therefore r = 1 \pm 2i $
$\implies y_c = c_1e^x\cos{2x} + c_2e^x\sin{2x}$
Next, solve non-homogenous equation using method of undetermined coefficients.
$\text{Consider:}~~~ y_p = A\sin{x} + B\cos{x}$
$\implies y'_p = A\cos{x} - B\sin{x}$
$\implies y''_p = -A\sin{x} -B\cos{x}$
Now, solve LHS.
$\begin {equation} \begin{array}{lcl}
y'' - 2y' + 5y &= -A\sin{x}-B\cos{x}-2(A\cos{x}-B\sin{x})+5(A\sin{x}+B\cos{x})
&= -A\sin{x} -B\cos{x} -2A\cos{x} + 2B\sin{x} + 5A\sin{x} + 5B\cos{x}
&= (2B+4A) \sin{x} + (4B-2A)\cos{x}
\end{array}
\end{equation}
$
$\implies 2B +4A = 1, ~~~ 4B-2A = 0$
$\implies A = \frac{1}{5}, ~~B = \frac{1}{10}$
$\therefore y_p = \frac{1}{5} \sin{x} + \frac{1}{10} \cos{x}$
Final solution is given by:
$y = y_p + y_c$
$\therefore y = c_1e^x\cos{2x} + c_2e^x\sin{2x} + \frac{1}{5}\sin{x} + \frac{1}{10}\cos{x}$
From the initial condition we find the constants $c_1$ and $c_2$:
$c_1=\frac{9}{10}, c_2=-\frac{1}{20}$
The solution is:
$y = \frac{9}{10}e^x\cos{2x} -\frac{1}{20}e^x\sin{2x} + \frac{1}{5}\sin{x} + \frac{1}{10}\cos{x}$
