Answer:
[tex]z(t) = Ae^{-9t} + B[/tex]
Step-by-step explanation:
z'' + 9z' = 0
z'' = -9z'
This is a second-order linear ordinary differential equation. Let z(t) has the form of
[tex]Ae^{kt} + B[/tex]
where A and B are constants and we are looking for constant k
The 1st derivative of z(t) with respect to t would be
[tex]z'(t) = Ake^{kt}[/tex]
And the 2nd derivative
[tex]z''(t) = Ak^2e^{kt}[/tex]
As z'' = -9z', we can substitute for z'' and z' and get:
[tex]Ak^2e^{kt} = -9Ake^{kt}[/tex]
[tex]k = -9[/tex]
Therefore [tex]z(t) = Ae^{-9t} + B[/tex]
