Linear Second Order ODE/y'' + y = exp -x cos x

Theorem
The second order ODE:
 * $(1): \quad y'' + y = e^{-x} \cos x$

has the general solution:
 * $y = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$

Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
 * $y'' + p y' + q y = \map R x$

where:
 * $p = 0$
 * $q = 1$
 * $\map R x = e^{-x} \cos x$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $y'' + y = 0$

From Linear Second Order ODE: $y'' + y = 0$, this has the general solution:
 * $y_g = C_1 \sin x + C_2 \cos x$

It remains to find a particular solution $y_p$ to $(1)$.

We have that:
 * $\map R x = e^{-x} \cos x$

From the Method of Undetermined Coefficients for Exponential of Sine and Cosine:
 * $y_p = e^{-x} \paren {A \cos x + B \sin x}$

where $A$ and $B$ are to be determined.

Hence:

Substituting into $(1)$:

Hence the result:
 * $y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x}$

So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:


 * $y = y_g + y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$

is the general solution to $(1)$.