Question

For what values of r does the function y = 6e^(rx) satsify the differential equation y'' - y' - 42y = 0

Answer 1

Compute the first derivative like this:
$(6e^{rx})'=6re^{rx}$
Second derivative:
$(6re^{rx})'=6r^2e^{rx}$
From the given equation we get:
$6r^2e^{rx}-6re^{rx}-42\times6e^{rx}=0$
Simplify like this:
$6r^2-6r-252=0$
Solving the above equation for r we get:
r=-6 or r=7