Linear Second Order ODE/y'' - y = 0/Proof 1

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Theorem

The second order ODE:

$(1): \quad y' ' - y = 0$

has the general solution:

$y = C_1 e^x + C_2 e^{-x}$


Proof

Note that:

\(\ds y_1\) \(=\) \(\ds e^x\)
\(\ds \leadsto \ \ \) \(\ds y'\) \(=\) \(\ds e^x\) Derivative of Exponential Function
\(\ds \leadsto \ \ \) \(\ds y' '\) \(=\) \(\ds e^x\) Derivative of Exponential Function

and so by inspection:

$y_1 = e^x$

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 $v$
\(\ds \) \(=\) \(\ds \int \dfrac 1 {e^{2 x} } C \rd x\)
\(\ds \) \(=\) \(\ds \int C e^{-2 x} \rd x\)
\(\ds \) \(=\) \(\ds -\frac {C e^{-2 x} } 2\) Primitive of $e^{a x}$


and so:

\(\ds y_2\) \(=\) \(\ds v y_1\) Definition of $y_2$
\(\ds \) \(=\) \(\ds e^x \paren {-\frac C 2 e^{-2 x} }\)
\(\ds \) \(=\) \(\ds -\frac C 2 e^{-x}\)


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

$y = C_1 \sin x + k \paren {-\dfrac C 2 e^{-x} }$

where $k$ is arbitrary.

Setting $C_2 = - \dfrac {k C} 2$ yields the result:

$y = C_1 e^x + C_2 e^{-x}$

$\blacksquare$


Sources

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