Linear Second Order ODE/y'' + 4 y = 6 cosine x
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Theorem
The second order ODE:
- $(1): \quad y + 4 y = 6 \cos x$
has the general solution:
- $y = C_1 \sin 2 x + C_2 \cos 2 x + 2 \cos x$
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:
- $p = 0$
- $q = 4$
- $\map R x = 6 \cos x$
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 $\cos x$ is not a particular solution of $(2)$.
So from the Method of Undetermined Coefficients for Sine and Cosine:
- $y_p = A \sin x + B \cos x$
where $A$ and $B$ are to be determined.
Hence:
\(\ds y_p\) | \(=\) | \(\ds A \sin x + B \cos x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}'\) | \(=\) | \(\ds A \cos x - B \sin x\) | Derivative of Sine Function, Derivative of Cosine Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds {y_p}\) | \(=\) | \(\ds -A \sin x - B \cos x\) | Derivative of Sine Function, Derivative of Cosine Function |
Substituting into $(1)$:
\(\ds -A \sin x - B \cos x + 4 \paren {A \sin x + B \cos x}\) | \(=\) | \(\ds 6 \cos x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds 3 A \sin x\) | \(=\) | \(\ds 0\) | equating coefficients | ||||||||||
\(\ds 3 B \cos x\) | \(=\) | \(\ds 6 \cos x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds A\) | \(=\) | \(\ds 0\) | |||||||||||
\(\ds B\) | \(=\) | \(\ds 2\) |
So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution:
- $y = y_g + y_p = C_1 \sin k x + C_2 \cos k x + 2 \cos x$
is the general solution to $(1)$.
$\blacksquare$