Linear Second Order ODE/y'' + k^2 y = sine b x
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:
- $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + \dfrac {\sin b x} {k^2 - b^2}$
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.18$: Problem $2$