Answer:
[tex]y = A e^{5t} + B e^{6t}[/tex]
Step-by-step explanation:
given,
[tex]y = e^{5t}[/tex]
and equation
y" - 11 y' + k y = 0...............(1)
now,
[tex]y' = 5 e^{5t}[/tex]
[tex]y"= 25 e^{5t}[/tex]
Putting value in equation (1)
[tex]25 e^{5t} - 11(5 e^{5t}) + k e^{5t} = 0[/tex]
[tex]- 30 + k = 0[/tex]
k = 30
now, differential equation becomes
y" - 11 y' +30 y = 0
( D² - 11 D + 30) y = 0
writing Auxiliary equation'
m² - 11 m + 30 = 0
(m - 5)(m-6) = 0
m = 5,6
now,
[tex]y = A e^{5t} + B e^{6t}[/tex]