Answer:
the value of k is 6 and 12.
Step-by-step explanation:
The differential equation is y" – 18y' + 72y = 0.
A solution of this differential equation is
[tex]y(x)=e^{kt}[/tex]
The first derivative is
[tex]y'(x)=ke^{kt}[/tex]
The second derivative is
[tex]y''(x)=k^2e^{kt}[/tex]
Substituting these values in the given DE
[tex]k^2e^{kt}-18ke^{kt}+72e^{kt}=0[/tex]
Factor out the GCF
[tex]e^{kt}(-k^2-18k+72)=0[/tex]
The function [tex]e^{kt}[/tex] can never be zero. Hence, we have
[tex]k^2-18k+72=0\\\\k^2-12k-6k+72=0\\\\k(k-12)-6(k-12)=0\\\\(k-12)(k-6)=0\\\\k=6,12[/tex]
Therefore, the value of k is 6 and 12.
Smaller value = 6
Larger value = 12
