Linear Second Order ODE/y'' - 2 y' + y = 2 x/Proof 1

Proof
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:
 * $y'' + p y' + q y = \map R x$

where:
 * $p = -2$
 * $q = 1$
 * $\map R x = 2 x$

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

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

It remains to find a particular solution $y_p$ to $(1)$.

Taking the expression $\map R x = 2 x$ and differentiating twice $x$:

Trying out $y = 2 x$ in $(1)$:
 * $0 - 4 + 2 x = 2 x - 4$

which is off by that constant of $4$.

But by Derivative of Constant:
 * $4' = 0$

Hence setting $y = 2 x + 4$:
 * $0 - 4 + \paren {2 x + 4} = 2 x$

and it can be seen that $y_p = 2 x + 4$ is indeed a particular solution to $(1)$.

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


 * $y = y_g + y_p = C_1 e^x + C_2 x e^x + 2 x + 4$

is the general solution to $(1)$.