Linear Second Order ODE/y'' + 10 y' + 25 y = 14 exp -5 x
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Theorem
The second order ODE:
- $(1): \quad y'' + 10 y' + 25 y = 14 e^{-5 x}$
has the general solution:
- $y = C_1 \cos 2 x + C_2 \sin 2 x + \sin x$
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:
- $p = 10$
- $q = 25$
- $\map R x = 14 e^{-5 x}$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $(2): \quad y'' + 10 y' + 25 y = 0$
From Linear Second Order ODE: $y'' + 10 y' + 25 y = 0$, this has the general solution:
- $y_g = C_1 e^{-5 x} + C_2 x e^{-5 x}$
We have that:
- $\map R x = 14 e^{-5 x}$
and it is noted that $14 e^{-5 x}$ is a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for the Exponential function:
- $y_p = A x^2 e^{-5 x}$
where:
- $A = \dfrac {14} 2$
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.18$: Problem $1 \ \text{(c)}$