Question

Quick 2nd order DE question http://i.imgur.com/8dToV.jpg

Answer 1

Ok so I found the roots, it's just the end I don't understand of the question, does it want to want me to just sub in my values in the general solution formula?

Answer 2

wait peice of cake
Roots are a+ib ,a-ib
solutions are c1*e^(a+ib)+c2*e^(a-ib)

Answer 3

Yeah so y=c1*e^(a+ib)+c2*e^(a-ib) that's the general solution

Answer 4

But the next question asks for the real general solution. This is confusing, let me post the question in its entirely

Answer 5

http://i.imgur.com/cpdFA.jpg

Answer 6

Equation 4 is the orginal equation by the way.

Answer 7

Put t=0 and find a relation in c1 and c2
find derivative put t=0 and find relation in c1 and c2
solve simultaneously

Answer 8

Is that how to do part 1? or 2 or 3

Answer 9

part 3 the last post

Answer 10

Part 1 and 2 first post

Answer 11

Ok so part 1 is want the roots
Part is is wanting the roots in y=c1*e^(a+ib)+c2*e^(a-ib)
and part 3 is finding c1 and c2?

Answer 12

I found the roots of the equation but its not in the form of \[e^{lamba t} \]