Entered | Answer Preview | Result | Messages |
---|---|---|---|
(r^2)*[e^(r*x)]-3*r*[e^(r*x)]-40*[e^(r*x)] | r^2*e^(rx)-3*r*e^(rx)-40*e^(rx) | correct | |
(e^rx)*[(r-8)*(r+5)] | | incorrect | Powers must be either 0 or 1 |
-5, 8 | −5,8 | correct |
Hello. What are we going to find to help you?
And great answer in the negative 1 and negative 2 question.
i mea a and c are correct but when i simplify the b as (e^rx)*[(r-8)*(r+5)] i gotta do something else i think
$$\\y"-3y'-40y=0.\\\\
Try \\
y=e^{rx}\qquad y'=re^{rx} \qquad y"=r^2e^{rx}\\\\
\mbox{The question becomes}\\\\
r^2e^{rx}-3re^{rx}-40e^{rx}=0\\\\
e^{rx}(r^2-3r-40)=0\\\\
e^{rx}(r-8)(r+5)=0\\\\
\mbox{The only solutions to this are }
r=8\;\; and\;\; r=-5$$
Dragon, this stuff is way over the top of your head. Sorry mate.
Milkshake,
Does that help?
But great answer, though. Keep it up, or lose the grand war of Camelot.