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

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

has the general solution:
 * $y = C_1 e^x + C_2 x e^x + \dfrac {x^2 e^x} 2$

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

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

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

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

and so from the Method of Undetermined Coefficients for the Exponential function:
 * $y_p = \dfrac {K x^2 e^{a x} } 2$

where:
 * $K = 1$
 * $a = 1$

Hence:

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


 * $y = y_g + y_p = C_1 e^x + C_2 x e^x + \dfrac {x^2 e^x} 2$