Linear Second Order ODE/y'' + 2 y' + 2 y = 0
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Theorem
The second order ODE:
- $(1): \quad y + 2 y' + 2 y = 0$
has the general solution:
- $y = e^{-x} \paren {A \cos x + B \sin x}$
Proof
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
- $(2): \quad: m^2 + 2 m + 2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
- $m_1 = -1 + i$
- $m_2 = -1 - i$
These are complex conjugates.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
- $y = e^{-x} \paren {A \cos x + B \sin x}$
$\blacksquare$
Verification
The equation:
- $(1): \quad y = e^{-x} \paren {A \cos x + B \sin x}$
is a set of solutions to the second order ODE:
- $y + 2 y' + 2 y = 0$
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 $1$