Question

give me a hint on how to solve the differential equation: -b*x'(t)+m*g=m*x''(t)

Answer 1

with x'(0)=0, x''(0)=0

Answer 2

I would re-write your -b*x'(t)+m*g=m*x''(t) in a more standard form:

m*x''(t)+bx'(t) - mg= 0. Before we do anything further, tell me whether or not you agree with this change of appearance.

Answer 3

Yea, i've done the same, my problem is when im trying to find the roots

Answer 4

You'll only be able to find roots as algebraic expressions, since no numerical values are given.

I've suggested that you focus on the d. e. m*x''(t)+bx'(t) - mg= 0. Upon closer examination, I see no x in there, only x'' and x'.

Are you certain that the last term (-mg) is not actually -mgx?

If it were, then we could write the auxiliary equation mp^2+bp +mg = 0. Get my drift?
I need a refresher myself, so am going to look up "auxiliary equation" on the 'Net.

Answer 5

My problem was that i couldn't use the auxiliary equation since i had no numerical values.

Answer 6

Please see:

http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf

and note how the d. e. includes x'', x' and x.
Your d. e. includes only x'' and x', not x.

Any idea of how to get around that omission?

Answer 7

the equation is given as -b*dx/dt+m*g=m*d^2x/dt^2
And is a differential equation for a falling object with air resistance -b

Answer 8

i can't write it any better

Answer 9

since latex isn't working

Answer 10

Actually, I don't see any problem in finding an aux. eq'n. I'd move the constant mg to the right side, leaving mx'' + bx' on the left side,

with the result that the aux. eq'n would be mp^2 + bp + 0=0,
which would factor to p(mp+b)=0, and result in roots p=0 and p=-b/m.

Answer 11

Does this make sense to you?
H-Z, I've been on OpenStudy by choice for about four hours straight, so pls understand if I excuse myself pretty darn quickly.

Answer 12

Yes i do understand and im very gratefull that you're helping me. I'll just try to work on it a bit for myself and I'll be back if it doesnt work out.

Answer 13

Great. Good luck! Hope to talk with yo u again soon. Bye.

Answer 14

see ya, and thanks :)

Answer 15

i figured it out :)