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


Hence:

\(\ds y_p\) \(=\) \(\ds A \cos 3 x + B \sin 3 x\)
\(\ds \leadsto \ \ \) \(\ds {y_p}'\) \(=\) \(\ds -3 A \sin 3 x + 3 B \cos 3 x\) Derivative of Sine Function, Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds {y_p}\) \(=\) \(\ds -9 A \cos 3 x - 9 B \sin 3 x\) Power Rule for Derivatives


Substituting into $(1)$:

\(\ds -9 A \cos 3 x - 9 B \sin 3 x - 2 \paren {-3 A \sin 3 x + 3 B \cos 3 x} - 5 \paren {A \cos 3 x + B \sin 3 x}\) \(=\) \(\ds 2 \cos 3 x - \sin 3 x\)
\(\ds \leadsto \ \ \) \(\ds -9 B \sin 3 x + 6 A \sin 3 x - 5 B \sin 3 x\) \(=\) \(\ds \sin 3 x\) equating coefficients
\(\ds -9 A \cos 3 x - 6 B \cos 3 x - 5 A \cos 3 x\) \(=\) \(\ds 2 \cos 3 x\)
\(\ds \leadsto \ \ \) \(\ds 6 A - 14 B\) \(=\) \(\ds 1\)
\(\ds -6 B - 14 A\) \(=\) \(\ds 2\)
\(\ds \leadsto \ \ \) \(\ds \paren {14^2 + 6^2} A\) \(=\) \(\ds -28 - 6\)
\(\ds \paren {14^2 + 6^2} B\) \(=\) \(\ds -12 + 14\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds -\dfrac {17} {116}\)
\(\ds B\) \(=\) \(\ds \dfrac 1 {116}\)


Hence the result:

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

$\blacksquare$


Sources

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