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mgier001
Oct25-11, 09:43 PM
My professor assigned us a problem where
m=16 lb
k=10 lb/ft
a=2
t(0)=1 ft
t'(0)=0 ft/sec

Through this I determined the DE
x''(t)+4x'(t)+20x(t)=e-t
with initial conditions mentioned above.

Using the method of m2+4m+20=0
I got roots of -2±4i
Therefore, my yc=e-2t[c1sin(4t)+c2cos(4t)]

Next I attempted to solve further the problem, but I did not understand what to do as the examples I could find throughout the internet and my textbook all are regarding F(t)=cos(t) or similar.

Any help or guidance would be appreciated. Unfortunately the problem is due tomorrow morning so any quick help is more preferred ;)

Thanks :smile:,
Matt!

mgier001
Oct25-11, 10:31 PM
I managed to find
c1=1/2
c2=1

So
yc=e-2t((1/2)sin4t+cos4t)

Now what to do :confused:

mgier001
Oct25-11, 11:19 PM
I did some more work, can someone verify that this is correct please? :)

http://i40.tinypic.com/21mg29f.jpg

Thanks.

mgier001
Oct25-11, 11:28 PM
Also, the graph I got from my general solution is provided by this link: http://www.wolframalpha.com/input/pdfGet.jsp?id=MSP27419hh14h12d30f86a00004d37db5hb5 g755fg&s=16&i=graph+y%3D%28e^%28-2t%29%29%28%281%2F2%29sin%284t%29%2Bcos%284t%29%29 %2B%28%2821%2F4%29e^%28-t%29%29

mgier001
Oct26-11, 09:48 PM
I actually ended up getting y=(e^(-2x))((8/17)sin(4x)+(16/17)cos(4x))+(1/17)e^(-x)
I hope this is right :).