Linear Second Order ODE/y'' - y = 0
Theorem
The second order ODE:
- $(1): \quad y - y = 0$
has the general solution:
- $y = C_1 e^x + C_2 e^{-x}$
Proof 1
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:
- $y = C_1 e^x + C_2 e^{-x}$
$\blacksquare$
Proof 2
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE.
Its auxiliary equation is:
- $(2): \quad: m^2 - 1 = 0$
From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are:
- $m_1 = 1$
- $m_2 = -1$
These are real and unequal.
So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is:
- $y = C_1 e^x + C_2 e^{- x}$
$\blacksquare$
Proof 3
This is an instance of:
which yields:
- $y = C_1 e^{k x} + C_2 e^{-k x}$
where $k = 1$.
Hence the result.
$\blacksquare$