Linear Second Order ODE/y'' + 2 b y' + a^2 y = K cosine omega x/b less than a

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Theorem

The second order ODE:

$(1): \quad y + 2 b y' + a^2 y = K \cos \omega x$ where $b^2 < a^2$

has the general solution:

$y = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$

where:

$\alpha = \sqrt {a^2 - b^2}$
$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$


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:

$p = 2 b$
$q = a^2$
$\map R x = K \cos \omega x$


First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:

$y + 2 b y' + a^2 y = 0$

From Linear Second Order ODE: $y + 2 b y' + a^2 y = 0: b < a$, this has the general solution:

$y_g = e^{-b x} \paren {C_1 \, \map \cos {\sqrt {a^2 - b^2} } x + C_2 \, \map \sin {\sqrt {a^2 - b^2} } x}$

Substituting $\alpha = \sqrt {a^2 - b^2}$, this can be written more compactly as:

$y_g = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x}$


We have that:

$\map R x = K \cos \omega x$


It is noted that $K \cos \omega x$ is not itself a particular solution of $(2)$.

So from the Method of Undetermined Coefficients for Sine and Cosine:

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

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


Hence:

\(\ds y_p\) \(=\) \(\ds A \sin \omega x + B \cos \omega x\)
\(\ds \leadsto \ \ \) \(\ds {y_p}'\) \(=\) \(\ds \omega A \cos \omega x - \omega B \sin \omega x\) Derivative of Sine Function, Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds {y_p}\) \(=\) \(\ds -\omega^2 A \sin \omega x - \omega^2 B \cos \omega x\) Power Rule for Derivatives


Substituting into $(1)$:

\(\ds -\omega^2 \paren {A \sin \omega x + B \cos \omega x} + 2 b \omega \paren {A \cos \omega x - B \sin \omega x} + a^2 \paren {A \sin \omega x + B \cos \omega x}\) \(=\) \(\ds K \cos \omega x\)
\(\ds \leadsto \ \ \) \(\ds -\omega^2 A \sin \omega x - 2 b \omega B \sin \omega x + a^2 A \sin \omega x\) \(=\) \(\ds 0\) equating coefficients
\(\ds -\omega^2 B \cos \omega x + 2 b \omega A \cos \omega x + a^2 B \cos \omega x\) \(=\) \(\ds K \cos \omega x\)
\(\ds \leadsto \ \ \) \(\ds \paren {a^2 - \omega^2} A - 2 b \omega B\) \(=\) \(\ds 0\)
\(\ds \paren {a^2 - \omega^2} B + 2 b \omega A\) \(=\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds \frac {2 b \omega B} {a^2 - \omega^2}\) \(=\) \(\ds A\)
\(\ds \leadsto \ \ \) \(\ds \paren {a^2 - \omega^2} B + 2 b \omega \frac {2 b \omega B} {a^2 - \omega^2}\) \(=\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds B \paren {\frac {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} {a^2 - \omega^2} }\) \(=\) \(\ds K\)
\(\ds \leadsto \ \ \) \(\ds K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}\) \(=\) \(\ds B\)
\(\ds \leadsto \ \ \) \(\ds \frac {2 b \omega} {a^2 - \omega^2} K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}\) \(=\) \(\ds A\)
\(\ds \leadsto \ \ \) \(\ds K \frac {2 b \omega} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2}\) \(=\) \(\ds A\)


Hence:

\(\ds y_p\) \(=\) \(\ds K \frac {2 b \omega} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \sin \omega x + K \frac {a^2 - \omega^2} {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \cos \omega x\)
\(\ds \) \(=\) \(\ds \frac K {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \paren {2 b \omega \sin \omega x + \paren {a^2 - \omega^2} \cos \omega x}\)


From Multiple of Sine plus Multiple of Cosine: Cosine Form:

\(\ds \) \(\) \(\ds 2 b \omega \sin \omega x + \paren {a^2 - \omega^2} \cos \omega x\)
\(\ds \) \(=\) \(\ds \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} \map \cos {\omega x + \map \arctan {\dfrac {-2 b \omega} {a^2 - \omega^2} } }\)
\(\ds \) \(=\) \(\ds \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} \map \cos {\omega x - \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} } }\) Tangent Function is Odd


and so:

\(\ds y_p\) \(=\) \(\ds \frac {K \sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } {\paren {a^2 - \omega^2}^2 + 4 b^2 \omega^2} \map \cos {\omega x - \phi}\) where $\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$
\(\ds \) \(=\) \(\ds \frac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}\)


So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:

$y = y_g + y_p = e^{-b x} \paren {C_1 \cos \alpha x + C_2 \sin \alpha x} + \dfrac K {\sqrt {4 b^2 \omega^2 + \paren {a^2 - \omega^2}^2} } \map \cos {\omega x - \phi}$

where:

$\phi = \map \arctan {\dfrac {2 b \omega} {a^2 - \omega^2} }$

$\blacksquare$

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