Linear Second Order ODE/y'' + y = 0/Proof 4

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Theorem

The second order ODE:

$(1): \quad y + y = 0$

has the general solution:

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


Proof

Note that:

\(\ds y_1\) \(=\) \(\ds \sin x\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds \cos x\) Derivative of Sine Function
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds -\sin x\) Derivative of Cosine Function

and so:

$y_1 = x$

is a particular solution of $(1)$.


$(1)$ is in the form:

$y + \map P x y' + \map Q x y = 0$

where:

$\map P x = 0$
$\map Q x = 1$


From Particular Solution to Homogeneous Linear Second Order ODE gives rise to Another:

$\map {y_2} x = \map v x \, \map {y_1} x$

where:

$\ds v = \int \dfrac 1 { {y_1}^2} e^{-\int P \rd x} \rd x$

is also a particular solution of $(1)$.


We have that:

\(\ds \int P \rd x\) \(=\) \(\ds \int 0 \rd x\)
\(\ds \) \(=\) \(\ds k\) where $k$ is arbitrary
\(\ds \leadsto \ \ \) \(\ds e^{-\int P \rd x}\) \(=\) \(\ds e^{-k}\)
\(\ds \) \(=\) \(\ds C\) where $C$ is also arbitrary


Hence:

\(\ds v\) \(=\) \(\ds \int \dfrac 1 { {y_1}^2} e^{- \int P \rd x} \rd x\) Definition of $v$
\(\ds \) \(=\) \(\ds \int \dfrac 1 {\sin^2 x} C \rd x\)
\(\ds \) \(=\) \(\ds \int C \csc^2 x \rd x\) Definition of Cosecant
\(\ds \) \(=\) \(\ds -C \cot x\) Primitive of $\csc^2 x$


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds \sin x \paren {-C \cot x}\)
\(\ds \) \(=\) \(\ds -C \cos x\) Definition of Cotangent, and simplifying


From Two Linearly Independent Solutions of Homogeneous Linear Second Order ODE generate General Solution:

$y = C_1 \sin x + k \paren {-C \cos x}$

where $k$ is arbitrary.


Setting $C_2 = - k C$ yields the result:

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

$\blacksquare$


Sources

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