Find the two values of k for which y(x) = ekt is a solution of the differential equation y" – 18y' + 72y = 0. smaller value = larger value = Preview Preview

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 = 6Larger value = 12