Linear Second Order ODE/y'' + 10 y' + 25 y = 14 exp -5 x

Theorem
The second order ODE:
 * $(1): \quad y'' + 10 y' + 25 y = 14 e^{-5 x}$

has the general solution:
 * $y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$

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:
 * $p = 10$
 * $q = 25$
 * $\map R x = 14 e^{-5 x}$

First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
 * $(2): \quad y'' + 10 y' + 25 y = 0$

From Linear Second Order ODE: $y'' + 10 y' + 25 y = 0$, this has the general solution:
 * $y_g = C_1 e^{-5 x} + C_2 x e^{-5 x}$

We have that:
 * $\map R x = 14 e^{-5 x}$

and it is noted that $14 e^{-5 x}$ is a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for the Exponential function:
 * $y_p = A x^2 e^{-5 x}$

where:
 * $A = \dfrac {14} 2$

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


 * $y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$