Question

If y = e^5t is a solution to the differential equation d^2 y/dt^2 - 13 dy/dt + ky = 0, find the value of the constant k and the general form y = Ae^5t + Be^at of the solution to the above equation, (i.e. find a). (Use constants A, B, etc., for any constants in your solution formula.)

Answer 1

k = -12/5

A = 125/12

B = -325/12

a = 5

Explanation:

y = e^5t

Dy/dt = 5e^5t

d2y/dt2 = 25e^5t

Inputting the values of dy/dt and d2y/dt2 into the equation above, we have:

25e^5t - 13e^5t + 5k(e^5t) = 0

12e^5t + 5k(e^5t) = 0

e^5t(12 + 5k) = 0

12 + 5k = 0

k = -12/5

The equation becomes,

d2y/dt2 - 13dy/dt -12/5y = o

So rearranging the equation,

-5/12d2y/dt2 + 65/12dy/dt + y = 0

y = 5/12(25e^5t) - 65/12(5e^5t)

y = 125/12e^5t - 325/12e^5t

Therefore,

k = -12/5

A = 125/12

B = -325/12

a = 5