Linear Second Order ODE/y'' + 4 y = 4 cosine 2 x
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Theorem
The second order ODE:
- $(1): \quad y + 4 y = 4 \cos 2 x$
has the general solution:
- $y = C_1 \sin 2 x + C_2 \cos 2 x + x \sin 2 x$
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:
- $p = 0$
- $q = 4$
- $\map R x = 4 \cos 2 x$
First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE:
- $(2): \quad y + 4 y = 0$
From Linear Second Order ODE: $y + 4 y = 0$, this has the general solution:
- $y_g = C_1 \sin 2 x + C_2 \cos 2 x$
It is noted that $4 \cos 2 x$ is a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
- $y_p = A x \sin 2 x + B x \cos 2 x$
where $A$ and $B$ are to be determined.
Hence:
| \(\ds y_p\) | \(=\) | \(\ds A x \sin 2 x + B x \cos 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds 2 A x \cos 2 x - 2 B x \sin 2 x + A \sin 2 x + B \cos 2 x\) | Derivative of Sine Function, Derivative of Cosine Function, Product Rule for Derivatives | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds {y_p}\) | \(=\) | \(\ds -4 A x \sin 2 x - 4 B x \cos 2 x + 2 A \cos 2 x - 2 B \sin 2 x + 2 A \cos 2 x - 2 B \sin 2 x\) | Derivative of Sine Function, Derivative of Cosine Function, Product Rule for Derivatives | ||||||||||
| \(\ds \) | \(=\) | \(\ds -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x\) |
Substituting into $(1)$:
| \(\ds -4 A x \sin 2 x - 4 B x \cos 2 x + 4 A \cos 2 x - 4 B \sin 2 x + 4 \paren {A x \sin 2 x + B x \cos 2 x}\) | \(=\) | \(\ds 4 \cos 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A \paren {4 - 4} x \sin 2 x - 4 B \sin 2 x\) | \(=\) | \(\ds 0\) | equating coefficients | ||||||||||
| \(\ds B \paren {4 - 4} x \cos 2 x + 4 A \cos 2 x\) | \(=\) | \(\ds 4 \cos 2 x\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds - 4 B\) | \(=\) | \(\ds 0\) | |||||||||||
| \(\ds 4 A\) | \(=\) | \(\ds 4\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds 1\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds B\) | \(=\) | \(\ds 0\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + x \sin 2 x$
is the general solution to $(1)$.
$\blacksquare$