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:

$y = y_g + y_p = C_1 e^{-5 x} + C_2 x e^{-5 x} + 7 x^2 e^{-5 x}$

$\blacksquare$


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