Linear Second Order ODE/y'' - 2 y' - 5 y = 2 cos 3 x - sin 3 x/Particular Solution/Exponential Form

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Theorem

The second order ODE:

$(1): \quad y'' - 2 y' - 5 y = 2 \cos 3 x - \sin 3 x$

has a particular solution:

$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$


Proof

From Linear Second Order ODE: $y'' - 2 y' - 5 y = 0$, we have established that the general solution to $(1)$ is:

$y_g = C_1 \, \map \exp {\paren {1 + \sqrt 6} x} + C_2 \, \map \exp {\paren {1 - \sqrt 6} x}$


We note that $2 \cos 3 x - \sin 3 x$ is not itself a particular solution of $(2)$.


From the Method of Undetermined Coefficients for Sine and Cosine:

$y_p = A \cos 3 x + B \sin 3 x$

where $A$ and $B$ are to be determined.

The right hand side of $(1)$ is the real part of $\paren {2 + i} e^{3 i x}$.


Thus, to find a particular solution of $(1)$ when the right hand side is $\paren {2 + i} e^{3 i x}$, we substitute $y = A e^{3 i x}$.

Hence:

\(\ds A \paren {\paren {3 i}^2 - 2 \paren {3 i} - 5}\) \(=\) \(\ds 2 + i\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds -\dfrac {2 + i} {14 + 6 i}\)
\(\ds \) \(=\) \(\ds -\dfrac {\paren {2 + i} \paren {14 - 6 i} } {14^2 + 6^2}\)
\(\ds \) \(=\) \(\ds -\dfrac {34 + 2 i} {232}\)
\(\ds \) \(=\) \(\ds -\dfrac 1 {116} \paren {17 + i}\)


Taking the real part of the corresponding solution $A e^{3 i x}$, we get:

\(\ds y\) \(=\) \(\ds \map \Re {-\dfrac 1 {116} \paren {17 + i} \paren {\cos 3 x + i \sin 3 x} }\)
\(\ds \) \(=\) \(\ds -\dfrac 1 {116} \paren {17 \cos 3 x - \sin 3 x}\)


Hence the result:

$y_p = \dfrac 1 {116} \paren {\sin 3 x - 17 \cos 3 x}$

$\blacksquare$


Sources