Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 1

Theorem

The second order ODE:

$(1): \quad y - y' - 6 y = e^{-x}$

has the general solution:

$y = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$

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:

$p = -1$

$q = -6$

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

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

$y - y' - 6 y = 0$

From Linear Second Order ODE: $y - y' - 6 y = 0$, this has the general solution:

$y_g = C_1 e^{3 x} + C_2 e^{-2 x}$

We have that:

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

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

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

where:

$K = 1$

$a = -1$

$p = -1$

$q = -6$

Hence:

\(\ds y_p\)

\(=\)

\(\ds \dfrac {e^{-x} } {1 + 1 - 6}\)

\(\ds \)

\(=\)

\(\ds -\dfrac {e^{-x} } 4\)

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

$y = y_g + y_p = C_1 e^{3 x} + C_2 e^{-2 x} - \dfrac {e^{-x} } 4$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.19$: Problem $2$