Question

If y = e 4t is a solution to the differential equation d 2 y dt2 −10 dy dt +ky = 0, find the value of the constant k and the general solution to this equation. k = y = (Use constants A, B, etc., for any constants in your solution

Answer 1

a. k = 24 b. y =
`Ae^(4t) + Be^(6t)`

Explanation:

a. Find the value of the constant k

Since
`y = e^(4t)` is a solution of d²y/dt² -10dy/dt + ky = 0, then
`y = e^(4t)` must satisfy the equation.

So, d²y/dt² =
`16e^(4t)` and dy/dt =
`4e^(4t)`

So, d²y/dt² -10dy/dt + ky = 0

`16e^(4t)` - 10(
`4e^(4t)`) + k(
`e^(4t)`) = 0

`16e^(4t)` -
`40e^(4t)` +
`ke^(4t)` = 0

-
`24e^(4t)` +
`ke^(4t)` = 0

-
`24e^(4t)` = -
`ke^(4t)`

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 =
`Ae^(4t) + Be^(6t)`