Find the general solution to the differential equation?

Question

s'' - 36s = 0
Give the solution in terms of the variable t. List the smaller root first
C1 = ?
C2 = ?

Answer 1

characteristic equation is: r^2-36 = 0

solving this gives: r = +/- 6

this gives the two solutions,

c1 e^(6t)

c2 e^(-6t)

either of these solutions work, because:

[the second derivative]-36*[the function itself] = 0.

Answer 2

Isn't it just C1 = c1 cosh 6t, C2 = c2 sinh 6t? The characteristic equation is s^2 - 36 = 0.

Could also be expressed as

C1 = c1 exp 6t, C2 = c2 exp(-6t),

equivalent to the above, but the first way allows you to be a little more smug.

General solution is s = c1 cosh 6t + c2 sinh 6t, for any constants c1 and c2.

Answer 3

s'' - 36s = 0

We could look at the as a system of two first order, linear, ODEs, or we could see that the answer has to be an exponential s = C(e^(at).

s'' - 36s = 0
a^2C(e^(at) - 36C(e^(at) = 0
a^2 - 36 = 0
a = ±6
The complete solution is a linear combination of C1e^(-6t) + C2e^(6t), for any C1 and C2. Those depend on initial conditions.

Answer 4

Let s = A*e^(c1*t)

Then

s" = c1^2*Ae^(c1*t) - 36 (A*e^(c1*t)) = 0

Thus:

c1^2 - 36 = 0 ---> c1 = +/- 6 ("+/-" sign comes from taking the square root).

So

s = Ae^(6*t) + Be^(-6*t)