Quote:
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Originally Posted by @$3.1415rin
As always with differential equations, we search a function x which fits in that equation above, where that relation is true. This is usually done by searching the solution of the homogenous DEq, in our case that's m x'' + k x' + d x = 0. You can solve this with an exponential approach, i.e. you set x = exp( l * t ), put it into the hom DEq and divide by exp ( l * t ). This way we get a quadratic expression for l : m l^2 + k l + d = 0 which can be easily solved : l = -k/2m +- Sqrt( k^2 - 4 d m )/2m
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I guess it may get confused Pierre a bit, so I'll try to explain it a bit more.
1.
(f(g(t)))'=(f'(g(t)))*g'(t)
2.
(exp(t))'=exp(t)
3.
(l*t)'=l(t)'=l
Well, when we know this we can try find x'' and x' from Aspirin's equation (I mean this
m x'' + k x' + d x = 0)
x'=(exp(l*t))'=(exp(l*t))*(l*t)'=(exp(l*t))*l=l*ex p(l*t)
x''=(x')'=(l*(exp(l*t))'=l*(exp(l*t))'=l*x'=l*l*(e xp(l*t))=(l^2)*exp(l*t)
so
m*x''=m*(l^2)*exp(l*t)
k*x'=k*l*exp(l*t)
d*x=d*exp(l*t)
and
m*(l^2)*exp(l*t)+k*l*exp(l*t)+d*exp(l*t)=0
then divide all equation by
exp(l*t)
finally
m*(l^2)+k*l+d=0
I hope now it will be more clear for Pierre. 
[EDIT]
Cleared a bit more - according to Aspirin's suggestion
[/EDIT]