Method of Undetermined Coefficients/Sine and Cosine/Particular Solution/i b is Root of Auxiliary Equation/Trigonometric Form

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Proof Technique

Consider the nonhomogeneous linear second order ODE with constant coefficients:

$(1): \quad y + b^2 y = \alpha \sin b x + \beta \cos b x$


The Method of Undetermined Coefficients can be used to find a particular solution to $(1)$ in the following manner.


Method and Proof

Let $\map {y_g} x$ be the general solution to:

$y + b^2 y = 0$

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

$\map {y_g} x + \map {y_p} x$

is the general solution to $(1)$.


It remains to find $\map {y_p} x$.


Assume that there is a particular solution to $(1)$ of the form:

$y_p = x \paren {A \sin b x + B \cos b x}$

We have:

\(\ds \frac {\d} {\d x} y_p\) \(=\) \(\ds x \paren {b A \cos b x - b B \sin b x} + \paren {A \sin b x + B \cos b x}\) Derivative of Sine Function, Derivative of Cosine Function, Product Rule for Derivatives
\(\ds \frac {\d^2} {\d x^2} y_p\) \(=\) \(\ds x \paren {-b^2 A \sin b x - b^2 B \cos b x} + \paren {b A \cos b x - b B \sin b x} + \paren {b A \cos b x - b B \sin b x}\) Derivative of Sine Function, Derivative of Cosine Function, Product Rule for Derivatives
\(\ds \) \(=\) \(\ds x \paren {-b^2 A \sin b x - b^2 B \cos b x} + 2 \paren {b A \cos b x - b B \sin b x}\)


Inserting into $(1)$:

\(\ds x \paren {-b^2 A \sin b x - b^2 B \cos b x} + 2 \paren {b A \cos b x - b B \sin b x} + b^2 x \paren {A \sin b x + B \cos b x}\) \(=\) \(\ds \alpha \sin b x + \beta \cos b x\)
\(\ds \leadsto \ \ \) \(\ds 2 \paren {b A \cos b x - b B \sin b x}\) \(=\) \(\ds \alpha \sin b x + \beta \cos b x\)
\(\ds \leadsto \ \ \) \(\ds 2 b A \cos b x\) \(=\) \(\ds \beta \cos b x\)
\(\ds \leadsto \ \ \) \(\ds -2 b B \sin b x\) \(=\) \(\ds \alpha \sin b x\)
\(\ds \leadsto \ \ \) \(\ds 2 b A\) \(=\) \(\ds \beta\)
\(\ds \leadsto \ \ \) \(\ds -2 b B\) \(=\) \(\ds \alpha\)


Hence $A$ and $B$ can be expressed in terms of $\alpha$ and $\beta$:

\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds \frac \beta {2 b}\)
\(\ds B\) \(=\) \(\ds -\frac \alpha {2 b}\)


Hence:

$y_p = \dfrac {\beta x \sin b x} {2 b} - \dfrac {\alpha x \cos b x} {2 b}$

$\blacksquare$


Sources

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