# Plotting differential equationsWatch

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#1
Hi,

2 questions:

(1) I'm comparing a plot of y'' + 5y' + 4y = 0, given y(0)=1 and y'(0)=0

[ thus y = -(1/3)exp(-4x) + (4/3)exp(-x) ]

to a plot of y'' + 2y' + 5y = 0, again, given y(0)=1 and y'(0)=0.

[ thus y = (1/2)exp(-x)sin(2x) + exp(-x)cos(2x) ]

Would it be correct to say that the latter differs from the former in that it has a damped sinusoidal component? I'm not sure how else to describe it.

(2) For the differential equation y'' + 5y' + 4y = 17sin(x), where y(0)=1 and y'(0)=0, I obtain an equation

y = -(2/3)exp(-4x) + (25/6)exp(-x) - (5/2)cos(x) +(3/2)sin(x)

I'm looking at the behaviour of the solution as x becomes large. Seems to me that there's nothing to note -- it just goes on oscillating. Am I missing something obvious?

Cheers!
0
13 years ago
#2
(1)
If you like jargon, the first equation is "over damped" and the second is "under damped".

http://en.wikipedia.org/wiki/Damping

(2)
You could say that the -(2/3)exp(-4x) + (25/6)exp(-x) part of the solution decays to 0, leaving just the oscillatory part.
0
#3
(Original post by Jonny W)
(1)
If you like jargon, the first equation is "over damped" and the second is "under damped".

http://en.wikipedia.org/wiki/Damping

(2)
You could say that the -(2/3)exp(-4x) + (25/6)exp(-x) part of the solution decays to 0, leaving just the oscillatory part.
Thanks Jonny W.
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