Linear Second Order ODE/y'' - 3 y' + 2 y = 14 sine 2 x - 18 cosine 2 x

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Theorem

The second order ODE:

$(1): \quad y - 3 y' + 2 y = 14 \sin 2 x - 18 \cos 2 x$

has the general solution:

$y = C_1 e^{3 x} + C_2 e^{-2 x} - 4 x e^{-2 x}$


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:

$p = -3$
$q = 2$
$\map R x = 14 \sin 2 x - 18 \cos 2 x$


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

$y - 3 y' + 2 y = 0$

From Linear Second Order ODE: $y - 3 y' + 2 y = 0$, this has the general solution:

$y_g = C_1 e^x + C_2 e^{2 x}$


We have that:

$\map R x = 14 \sin 2 x - 18 \cos 2 x$

and it is noted that $14 \sin 2 x - 18 \cos 2 x$ is not itself a particular solution of $(2)$.

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

$y_p = A \sin 2 x + B \cos 2 x$

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


Hence:

\(\ds y_p\) \(=\) \(\ds A \sin 2 x + B \cos 2 x\)
\(\ds \leadsto \ \ \) \(\ds {y_p}'\) \(=\) \(\ds 2 A \cos 2 x - 2 B \sin 2 x\) Derivative of Sine Function, Derivative of Cosine Function
\(\ds \leadsto \ \ \) \(\ds {y_p}\) \(=\) \(\ds -4 A \sin 2 x - 4 B \cos 2 x\) Power Rule for Derivatives


Substituting into $(1)$:

\(\ds -4 A \sin 2 x - 4 B \cos 2 x - 3 \paren {2 A \cos 2 x - 2 B \sin 2 x} + 2 \paren {A \sin 2 x + B \cos 2 x}\) \(=\) \(\ds 14 \sin 2 x - 18 \cos 2 x\)
\(\ds \leadsto \ \ \) \(\ds -4 A \sin 2 x + 6 B \sin 2 x + 2 A \sin 2 x\) \(=\) \(\ds 14 \sin 2 x\) equating coefficients
\(\ds -4 B \cos 2 x - 6 A \cos 2 x + 2 B \cos 2 x\) \(=\) \(\ds - 18 \cos 2 x\)
\(\ds \leadsto \ \ \) \(\ds -2 A + 6 B\) \(=\) \(\ds 14\)
\(\ds -2 B - 6 A\) \(=\) \(\ds -18\)
\(\ds \leadsto \ \ \) \(\ds -A + 3 B\) \(=\) \(\ds 7\)
\(\ds 3 A + B\) \(=\) \(\ds 9\)
\(\ds \leadsto \ \ \) \(\ds -A + 3 \paren {-3 A + 9}\) \(=\) \(\ds 7\)
\(\ds \leadsto \ \ \) \(\ds -10 A\) \(=\) \(\ds -20\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds 2\)
\(\ds \leadsto \ \ \) \(\ds -2 + 3 B\) \(=\) \(\ds 7\)
\(\ds \leadsto \ \ \) \(\ds B\) \(=\) \(\ds 3\)

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

$y = y_g + y_p = C_1 e^x + C_2 e^{2 x} + 2 \sin 2 x + 3 \cos 2 x$

$\blacksquare$


Sources

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