Linear Second Order ODE/y'' - y = 3 exp -x

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Theorem

The second order ODE:

$(1): \quad y'' - y = 3 e^{-x}$

has the general solution:

$y = C_1 e^x + C_2 e^{-x} - \dfrac {3 x e^{-x} } 2$


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:

$p = 0$
$q = -1$
$\map R x = 3 e^{-x}$


First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y'' - y = 0$

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

$y_g = C_1 e^x + C_2 e^{-x}$


We have that:

$\map R x = 3 e^{-x}$

and so from the Method of Undetermined Coefficients for the Exponential function:

$y_p = \dfrac {K x e^{a x} } {2 a + p}$

where:

$K = 3$
$a = -1$
$p = 0$


Hence:

\(\ds y_p\) \(=\) \(\ds \dfrac {3 x e^{-x} } {2 \times \paren {-1} + 0}\)
\(\ds \) \(=\) \(\ds \dfrac {3 x e^{-x} } {-2}\)
\(\ds \) \(=\) \(\ds -\dfrac {3 x e^{-x} } 2\)


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

$y = y_g + y_p = C_1 e^x + C_2 e^{-x} - \dfrac {3 x e^{-x} } 2$

$\blacksquare$


Sources