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

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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 $\map {y_p} x$.


Consider the auxiliary equation to $(1)$:

$(2): \quad m^2 + b^2 = 0$


We have that $i b$ is a root of $(2)$.


Trigonometric Form

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:

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


Exponential Form

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

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

From Euler's Formula:

$\cos b x + i \sin b x = e^{i b x}$

and so:

$x \paren {A \sin b x + B \cos b x}$ is the real part of $x \paren {A - i B} \paren {\cos b x + i \sin b x} = x \paren {A - i B} e^{i b x}$

It is assumed that $A$, $B$, $p$ and $q$ are all real numbers.


Suppose we have found a solution $y$ of $(1)$ where:

$\map f x = \map {f_1} x + i \, \map {f_2} x$

where $\map y x$ and $\map f x$ are complex-valued.

Letting $\map y x = \map {y_1} x + \map {y_2} x$, where $y_1$ and $y_2$ are the real and imaginary parts of $\map y x$, we have:

${y_1} + p {y_1}' + q y_1 + i \paren { {y_2} + p {y_2}' + q y_2} = \map {f_1} x + i \, \map {f_2} x$

Equating real parts:

${y_1} + p {y_1}' + q y_1 = \map {f_1} x$

Equating imaginary parts:

${y_2} + p {y_2}' + q y_2 = \map {f_2} x$


Thus if $y$ is a particular solution to $(1)$ when the right hand side is $\map f x$:

$\map \Re y$ is a particular solution to $(1)$ when the right hand side is $\map \Re {\map f x}$
$\map \Im y$ is a particular solution to $(1)$ when the right hand side is $\map \Im {\map f x}$

So to find a particular solution when the right hand side is $K \cos x$ or $K \sin x$, we can first find a particular solution when the right hand side is $K e^{i b x}$ and then take its real part or imaginary part as necessary.


Hence, when we have $A \cos b x + B \sin b x$ on the right hand side:

replace it with $x \paren {A - i B} e^{i b x}$
use the Method of Undetermined Coefficients for Exponential functions

and then take its real part.

$\blacksquare$


Sources

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