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

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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 Linear Second Order ODE: $y'' - 2 y' + y = 0$, this has the general solution:

$y_g = C_1 e^x + C_2 x e^x$


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:

\(\ds y_p\) \(=\) \(\ds \dfrac {x^2 e^x} 2\)


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

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

$\blacksquare$


Sources