The easiest way of finding $\displaystyle y_p$ when function on the right is a sum of trigonometric functions is to try

$\displaystyle y_p = A \sin x + B \cos x$

and equate coefficients.

I.e.

$\displaystyle y_{p}' = A \cos x - B \sin x$

$\displaystyle y_{p}'' = - (A \sin x + B \cos x) = - y_p$

so substituting for $\displaystyle y_{p}''$ in your equation gives

$\displaystyle y_{p}'' - y_{p} = -2 y_{p} = 2 \cos x - 2 \sin x$

$\displaystyle \Leftrightarrow -A \sin x - B \cos x = \cos x - \sin x$.

I.e. $\displaystyle A = 1$ and $\displaystyle B = -1$ .

So the final solution given by $\displaystyle y = y_c + y_p$ is

$\displaystyle \boxed{ y= c_1 e^{-x} + c_2 e^{x} + \sin x - \cos x }$ .