Linear Second Order ODE/y'' - y' - 6 y = exp -x/Proof 1
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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 $2$