Linear Second Order ODE/y'' + 2 y' + 2 y = 0/Verification

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Theorem

The equation:

$(1): \quad y = e^{-x} \paren {A \cos x + B \sin x}$

is a set of solutions to the second order ODE:

$y'' + 2 y' + 2 y = 0$


Proof

Differentiating $(1)$ twice with respect to $x$ gives:

\(\ds y'\) \(=\) \(\ds -e^{-x} \paren {A \cos x + B \sin x} + e^{-x} \paren {-A \sin x + B \cos x}\)
\(\ds \) \(=\) \(\ds e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x}\) rearranging
\(\ds \leadsto \ \ \) \(\ds y''\) \(=\) \(\ds -e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} + e^{-x} \paren {-\paren {B - A} \sin x - \paren {A + B} \cos x}\)
\(\ds \) \(=\) \(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x}\) rearranging


Then:

\(\ds y'' + 2 y' + 2 y\) \(=\) \(\ds e^{-x} \paren {-2 B \cos x + 2 A \sin x} + 2 \paren {e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} } + 2 e^{-x} \paren {A \cos x + B \sin x}\)
\(\ds \) \(=\) \(\ds e^{-x} \paren {\paren {-2 B + 2 \paren {B - A} + 2 A} \cos x - \paren {2 A - 2 \paren {A + B} + 2 B} \sin x}\) rearranging
\(\ds \) \(=\) \(\ds 0\) simplifying

thus demonstrating that $(1)$ is indeed a solution to the second order ODE given.

$\blacksquare$


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