Linear Second Order ODE/y'' - 3 y' + 2 y = 5 exp 3 x

Theorem
The second order ODE:
 * $(1): \quad y'' - 3 y' + 2 y = 5 e^{3 x}$

has the general solution:
 * $y = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$

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:
 * $p = -3$
 * $q = 2$
 * $\map R x = 5 e^{3 x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $y'' - 3 y' + 2 y = 0$

From Linear Second Order ODE: $y'' - 3 y' + 2 y = 0$, this has the general solution:
 * $y_g = C_1 e^x + C_2 e^{2 x}$

We have that:
 * $\map R x = 5 e^{3 x}$

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:
 * $K = 5$
 * $a = 3$
 * $p = -3$
 * $q = 2$

Hence:

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


 * $y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + \dfrac {5 e^{3 x} } 2$