Linear Second Order ODE/y'' - 2 y' = 12 x - 10

Theorem
The second order ODE:
 * $(1): \quad y'' - 2 y = 12 x - 10$

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

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:
 * $p = 2$
 * $q = 0$
 * $\map R x = 12 x - 10$

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

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

We have that:
 * $\map R x = 12 x - 10$

and it is noted that $12 x - 10$ is not itself a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for Polynomials:
 * $y_p = A_0 + A_1 x + A_2 x^2$

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

Hence:

Substituting into $(1)$:

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


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