Linear Second Order ODE/y'' - 2 y' + 5 y = 25 x^2 + 12

Theorem
The second order ODE:
 * $(1): \quad y'' - 2 y' + 5 y = 25 x^2 + 12$

has the general solution:
 * $y = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$

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:
 * $p = -2$
 * $q = 5$
 * $\map R x = 25 x^2 + 12$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' - 2 y' + 5 y = 0$

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

We have that:
 * $\map R x = 25 x^2 + 12$

and it is noted that $25 x^2 + 12$ is not itself a particular solution of $(2)$.

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

for $A_n$ to be determined.

Hence:

Substituting into $(1)$:

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


 * $y = y_g + y_p = e^x \paren {C_1 \cos 2 x + C_2 \sin 2 x} + 2 + 4 x + 5 x^2$