Linear Second Order ODE/y'' + k^2 y = sine b x

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Theorem

The second order ODE:

$(1): \quad y'' + k^2 y = \sin b x$

has the general solution:

$y = \begin{cases} C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2} & : b \ne k \\ C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k} & : b = k \end{cases}$


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:

$p = 0$
$q = k^2$
$\map R x = \sin b x$


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

$(2): \quad y'' + k^2 y = 0$

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

$y_g = C_1 \sin k x + C_2 \cos k x$


We have that:

$\map R x = \sin b x$

There are two cases to address:

$b = k$
$b \ne k$


First suppose that $b = k$.

It is noted that $\sin b x = \sin k x$ is a particular solution of $(2)$.

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

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

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


Hence:

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


Substituting into $(1)$:

\(\ds -A k^2 x \sin k x - B k^2 x \cos k x + 2 A k \cos k x - 2 B k \sin k x + k^2 \paren {A x \sin k x + B x \cos k x}\) \(=\) \(\ds \sin k x\)
\(\ds \leadsto \ \ \) \(\ds A \paren {k^2 - k^2} x \sin k x - 2 B k \sin k x\) \(=\) \(\ds \sin k x\) equating coefficients
\(\ds B \paren {k^2 - k^2} x \cos k x + 2 A k \cos k x\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds - 2 B k\) \(=\) \(\ds 1\)
\(\ds 2 A k\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds B\) \(=\) \(\ds -\frac 1 {2 k}\)


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

$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x - \dfrac {x \cos k x} {2 k}$

$\Box$


Now suppose that $b \ne k$.

It is noted that $\sin b x$ is not a particular solution of $(2)$.

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

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

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


Hence:

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


Substituting into $(1)$:

\(\ds -A b^2 \sin b x - B b^2 \cos b x + k \paren {A \sin b x + B \cos b x}\) \(=\) \(\ds \sin b x\)
\(\ds \leadsto \ \ \) \(\ds A \paren {k^2 - b^2} \sin b x\) \(=\) \(\ds \sin b x\) equating coefficients
\(\ds B \paren {k^2 - b^2} \cos b x\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds 0\)
\(\ds B\) \(=\) \(\ds \frac 1 {k^2 - b^2}\)


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

$y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$

$\blacksquare$


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