Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x

Theorem

The second order ODE:

$(1): \quad y - 2 y' - 5 y = 2 \cos 3 x - \sin 3 x$

has the general solution:

$y = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x} + \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$

Proof

It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form:

$y + p y' + q y = \map R x$

where:

$p = -2$

$q = -5$

$\map R x = 2 \cos 3 x - \sin 3 x$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y - 2 y' - 5 y = 0$

From Linear Second Order ODE: $y - 2 y' - 5 y = 0$, this has the general solution:

$y_g = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x}$

We have that:

$\map R x = 2 \cos 3 x - \sin 3 x$

and it is noted that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.

We then determine the particular solution:

Particular Solution

The second order ODE:

$(1): \quad y - 2 y' - 5 y = 2 \cos 3 x - \sin 3 x$

has a particular solution:

$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$

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

$y = y_g + y_p = C_1 \map \exp {\paren {1 + \sqrt 6} x} + C_2 \map \exp {\paren {1 - \sqrt 6} x} + \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$

$\blacksquare$