Linear Second Order ODE/y'' - 4 y' - 5 y = x^2/y(0) = 1, y'(0) = -1

Theorem
Consider the second order ODE:
 * $(1): \quad y'' - 4 y' - 5 y = x^2$

whose initial conditions are:
 * $y = 1$ when $x = 0$
 * $y' = -1$ when $x = 0$

$(1)$ has the particular solution:
 * $y = \dfrac {e^{5 x} } {375} + \dfrac {4 e^{-x} } 3 - \dfrac {x^2} 5 + \dfrac {8 x} {25} - \dfrac {42} {125}$

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:
 * $p = -4$
 * $q = -5$
 * $\map R x = x^2$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' - 4 y' - 5 y = 0$

From Second Order ODE: $y'' - 4 y' - 5 y = 0$, this has the general solution:
 * $y_g = C_1 e^{5 x} + C_2 e^{-x}$

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

and it is noted that $x^2$ 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:


 * $y = y_g + y_p = C_1 e^{5 x} + C_2 e^{-x} - \dfrac 1 5 x^2 + \dfrac 8 {25} x - \dfrac {42} {125}$

Differentiating $x$:


 * $y' = 5 C_1 e^{5 x} - C_2 e^{-x} - \dfrac 2 5 x + \dfrac 8 {25}$

Substituting the initial conditions, remembering that $e^0 = 1$:

The result follows.