Linear Second Order ODE/y'' + k^2 y = 0/Proof 2
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Theorem
The second order ODE:
- $(1): \quad y'' + k^2 y = 0$
has the general solution:
- $y = A \, \map \sin {k x + B}$
or can be expressed as:
- $y = C_1 \sin k x + C_2 \cos k x$
Proof
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
- $(2): \quad: m^2 + k^2 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
- $m_1 = k i$
- $m_2 = -k i$
These are complex and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
- $y = C_1 \cos k x + C_2 \sin k x$
or, by disposing the constants differently:
- $y = C_1 \sin k x + C_2 \cos k x$
$\blacksquare$