# Linearly Independent Solutions of y'' - y = 0

## Theorem

The second order ODE:

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

has solutions:

$y_1 = e^x$
$y_2 = e^{-x}$

which are linearly independent.

## Proof

We have that:

 $\ds \frac \d {\d x} \, e^x$ $=$ $\ds e^x$ Derivative of Exponential Function $\ds \frac \d {\d x} \, e^{-x}$ $=$ $\ds -e^{-x}$ Derivative of Exponential Function $\ds \leadsto \ \$ $\ds \frac {\d^2} {\d x^2} \, e^x$ $=$ $\ds e^x$ $\ds \frac {\d^2} {\d x^2} \, e^{-x}$ $=$ $\ds e^{-x}$

Hence it can be seen by inspection that:

 $\ds \map {y_1} x$ $=$ $\ds e^x$ $\ds \map {y_2} x$ $=$ $\ds e^{-x}$

are solutions to $(1)$.

Calculating the Wronskian of $y_1$ and $y_2$:

 $\ds \map W {y_1, y_2}$ $=$ $\ds \begin{vmatrix} e^x & e^{-x} \\ e^x & -e^{-x} \end{vmatrix}$ $\ds$ $=$ $\ds -e^x e^{-x} - e^x e^{-x}$ $\ds$ $=$ $\ds -2$

So the Wronskian of $y_1$ and $y_2$ is never zero.

$y_1$ and $y_2$ are linearly independent.

$\blacksquare$