Linear Second Order ODE/y'' - 7 y' - 5 y = x^3 - 1

Theorem
The second order ODE:
 * $(1): \quad y'' - 7 y' - 5 y = x^3 - 1$

has the general solution:
 * $y = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$

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:
 * $p = -7$
 * $q = -5$
 * $\map R x = x^3 - 1$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' - 7 y' - 5 y = 0$

From Linear Second Order ODE: $y'' - 7 y' - 5 y = 0$, this has the general solution:
 * $y_g = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x}$

We have that:
 * $\map R x = x^3 - 1$

and it is noted that $x^3 - 1$ is not itself a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for Polynomials:
 * $y_p = A_0 + A_1 x + A_2 x^2 + A_3 x^3$

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:


 * $y = y_g + y_p = C_1 \, \map \exp {\paren {\dfrac 7 2 + \dfrac {\sqrt {69} } 2} x} + C_2 \, \map \exp {\paren {\dfrac 7 2 - \dfrac {\sqrt {69} } 2} x} + \dfrac 1 {625} \paren {-125 x^3 + 525 x^2 - 1620 x + 2603}$