Linear Second Order ODE/y'' + 4 y' + 5 y = 2 exp -2 x

Theorem
The second order ODE:
 * $(1): \quad y'' + 4 y' + 5 y = 2 e^{-2 x}$

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

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:
 * $p = 4$
 * $q = 5$
 * $\map R x = 2 e^{-2 x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $y'' + 4 y' + 5 y = 0$

From Second Order ODE: $y'' + 4 y' + 5 y = 0$, this has the general solution:
 * $y_g = e^{-2 x} \paren {C_1 \cos x + C_2 \sin x}$

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

We have that:
 * $\map R x = 2 e^{-2 x}$

and so from the Method of Undetermined Coefficients for the Exponential Function:
 * $y_p = \dfrac {K e^{a x} } {a^2 + p a + q}$

where:
 * $K = 2$
 * $a = -2$
 * $p = 4$
 * $q = 5$

Hence:

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


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

is the general solution to $(1)$.