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.

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

$y_1$ and $y_2$ are linearly independent.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 3.15$: Problem $1$