Linearly Independent Solutions of y'' - y = 0

From ProofWiki
Jump to navigation Jump to search


The second order ODE:

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

has solutions:

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

which are linearly independent.


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.

Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:

$y_1$ and $y_2$ are linearly independent.