Answer:
a. k = 24 b. y = [tex]Ae^{4t} + Be^{6t}[/tex]
Step-by-step explanation:
a. Find the value of the constant k
Since [tex]y = e^{4t}[/tex] is a solution of d²y/dt² -10dy/dt + ky = 0, then [tex]y = e^{4t}[/tex] must satisfy the equation.
So, d²y/dt² = [tex]16e^{4t}[/tex] and dy/dt = [tex]4e^{4t}[/tex]
So, d²y/dt² -10dy/dt + ky = 0
[tex]16e^{4t}[/tex] - 10([tex]4e^{4t}[/tex]) + k([tex]e^{4t}[/tex]) = 0
[tex]16e^{4t}[/tex] - [tex]40e^{4t}[/tex] + [tex]ke^{4t}[/tex] = 0
- [tex]24e^{4t}[/tex] + [tex]ke^{4t}[/tex] = 0
- [tex]24e^{4t}[/tex] = -[tex]ke^{4t}[/tex]
So, k = 24
b. Find the general solution to this equation.
Since k = 24, our equation is
d²y/dt² -10dy/dt + 24y = 0
The characteristic equation is thus
m² - 10m + 24 = 0
Factorizing this we have
m² - 6m - 4m + 24 = 0
m(m - 6) - 4(m - 6) = 0
(m - 6)(m - 4) = 0
m - 6 = 0 or m - 4 = 0
m = 6 or m = 4
Since we have real roots for the characteristic equation, then, the general solution is
y = [tex]Ae^{4t} + Be^{6t}[/tex]
