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.
$\blacksquare$
Sources
- 1958: G.E.H. Reuter: Elementary Differential Equations & Operators ... (previous) ... (next): Chapter $1$: Linear Differential Equations with Constant Coefficients: $\S 2$. The second order equation: $\S 2.1$ The reduced equation: Example $1$