Help..can anybody solve this?

Question

Newton's second law :

m*a = P*s*e^(-bt) - C*v(t)

a = d^2x / dt

v = dx / dt

m*d^2x / dt = P*s*e^(-bt) - C*dx / dt

m*d^2x / dt + C*dx / dt = P*s*e^(-bt) >>> Differential equation

^how can i solve that differential equation?

Making the following assumptions:
Pressure outside can = p, mass of can = m, Area of can ends (each) = A, area of valve hole = s

If x(t) is the instantaneous position of the can, x' = dx/dt = v, and x'' = dv/dt = d^2x/dt^2 = a. Substitute into your equation.
mx'' = Pse^-bt - Cx', Next rearrange and get it into some sort of "standard form" I got
x'' + k1x' + k2e^-bt = 0 where k1=C/m and k2=(-Ps/m)
x(0) = 0, X'(0) = 0

will be in terms of C,b,m, P and s

Answer 1

I am not a physicist, but i can answer as a mathematician. Is m and c constant??
then mr^2 + cr = 0 is your characteristic equation.
factorise to get: r(mr+c) = 0
r = 0 or r = -c/m thus your homogeneous solution is:
x = L+ Re^(-ct/m)
Now we have to look for particular solution.
Your particular solution will take the form
x= Ve^(-bt) (provided P and s are constants, which I am not clear about. Thus
x' = -bVe^(-bt)
x''= b^2Ve^(-bt)
substitute into your original equation and solve for V.
m b^2Ve^(-bt) + C( -bVe^(-bt)) = P*s*e^(-bt)
hence
(mb^2)V - bCV = Ps (equating co-efficients)
and V = P*s/(mb^2 - bC)
and x is therefore the sum of particular integral and the homogeneous solution
x = L+ Re^(-ct/m) + P*s/(mb^2 - bC)e(-bt)

Note: x'' + k1x' + k2e^-bt = 0 is not standard form. standard form would be
ax'' + bx' + cx = f(t)

Answer 2

Hmmm.....
x'' + k1x' + k2e^-bt = 0
That's a bit of a nasty bastard. Without the exponential term it's a simple 2'nd order, 1'st degree differential equation, but the exponential term is going to require some serious analysis.

You might try looking around at some of the math sites (such as 'the electronic journal of differential equations') to see if someone has already got a cannonical solution, or if it's a 'standard form' equation.

On a brighter note, at least you have some fairly simple boundry values ☺

Doug

Answer 3

Wow, differential equations. Sorry but I also got no talent at this. Anyway good luck, hope somebody help us up.

Answer 4

u gotta ask in homework section, orelse, nobody else will notice this


hope that helps