using you knowledge of derivatives (particularly setting y=e^kt) solve the following equation:
y'' + 7y' +3y=0
- 1 decade agoFavorite Answer
the general equation is y(t)= Ae^(-7+sqr root37/2) + Be(-7-sqr root37/2). Where A and B are some arbitrary constants.
- mathsmanretiredLv 71 decade ago
I would just add a little further explanation. If you start with
y = constant*e^kt then as shown above you get two values of k. Call these k1 and k2. This means that y = constant*e^k1t and
y = constant*e^k2t are both possible solutions of the differential equation and the constants don't have to be the same. It can further be shown that these two added together are also a solution to the differential equation so the most general answer
is y = Ae^k1t + Be^k2t as given by the first person. (By the way, I think that helping a person find the answer themselves is far more important than just giving it to them.) Also this satisfies the general idea that a differential equation should contain the same number of arbitrary constants as the degree of the highest differential in it.
- ThermoLv 61 decade ago
Guess y = ae^bx
Then y ' = abe^bx and y"= ab^2e^bx
Poke this into y'' + 7y' +3y=0
ab^2e^bx + 7abe^bx + 3ae^bx = 0
This must be true for any x.
e^bx never equals 0, so
ab^2 + 7ab + 3a = 0
b^2 + 7b + 3 = 0 or a=0
If a=0 then b can be any
if b^2 + 7b + 3 = 0 then b = ... or ... and b can be any real number