Linear Second Order ODE/y'' + y = exp -x cos x

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Theorem

The second order ODE:

$(1): \quad y'' + y = e^{-x} \cos x$

has the general solution:

$y = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$


Proof

It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form:

$y'' + p y' + q y = \map R x$

where:

$p = 0$
$q = 1$
$\map R x = e^{-x} \cos x$


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

$y'' + y = 0$

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

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


It remains to find a particular solution $y_p$ to $(1)$.


We have that:

$\map R x = e^{-x} \cos x$

From the Method of Undetermined Coefficients for Exponential of Sine and Cosine:

$y_p = e^{-x} \paren {A \cos x + B \sin x}$

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


Hence:

\(\ds y_p\) \(=\) \(\ds e^{-x} \paren {A \cos x + B \sin x}\)
\(\ds \leadsto \ \ \) \(\ds {y_p}'\) \(=\) \(\ds -e^{-x} \paren {A \cos x + B \sin x} + e^{-x} \paren {-A \sin x + B \cos x}\) Product Rule for Derivatives etc.
\(\ds \) \(=\) \(\ds e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x}\)
\(\ds \leadsto \ \ \) \(\ds {y_p}''\) \(=\) \(\ds -e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} + e^{-x} \paren {-\paren {B - A} \sin x - \paren {A + B} \cos x}\) Product Rule for Derivatives etc.
\(\ds \) \(=\) \(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x}\)


Substituting into $(1)$:

\(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x} + e^{-x} \paren {A \cos x + B \sin x}\) \(=\) \(\ds e^{-x} \cos x\)
\(\ds \leadsto \ \ \) \(\ds -2 B \cos x + A \cos x\) \(=\) \(\ds \cos x\) equating coefficients
\(\ds 2 A \sin x + B \sin x\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds A - 2 B\) \(=\) \(\ds 1\)
\(\ds 2 A + B\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds 5 A\) \(=\) \(\ds 1\)
\(\ds 5 B\) \(=\) \(\ds -2\)
\(\ds \leadsto \ \ \) \(\ds A\) \(=\) \(\ds \dfrac 1 5\)
\(\ds B\) \(=\) \(\ds -\dfrac 2 5\)


Hence the result:

$y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x}$

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

$y = y_g + y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$

is the general solution to $(1)$.

$\blacksquare$


Sources