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

Answer:

a. k = 24 b. y = $Ae^{4t} + Be^{6t}$

Step-by-step 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 = $Ae^{4t} + Be^{6t}$