Solve the initial-value problem

maeveoneill

Junior Member
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Sep 24, 2005
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93
I have the completed this question, but the answer in the back of my textbook gives me a different asnwer, if someone can explain to me why or if that answer is incorrect.

Solve in the inital value problem:
y'' + 2y' + 2y =0, y(0) =2, y'(0)=1

The auxillary equations is r^2 + 2r + 2= 0, whose roots are -1+i, -1-i
-> fish symbol thing =-1, B= 1

y(x) = e^-x (c1cosx + c2sinx)
y(o)=c1= 2, y'(O)= c2=1

y(X) = e^-x (2cosx + sinx) --- my solution, however the back of my textbook has the constatn 3 infront of sinx ... how do you get that answer if it is correct??[/img][/list]
 
Here's it is using LaPlace Transforms.

\(\displaystyle \L\\y''+2y'+2y=0, \;\ y(0)=2, \;\ y'(0)=1\)

\(\displaystyle \L\\p^{2}Y-py(0)-y'(0)+2(pY-y(0))+2Y=0\)

\(\displaystyle \L\\p^{2}Y-2p-1+2pY-4+2Y=0\)

Solve for Y:

\(\displaystyle \L\\Y=\frac{5+2p}{p^{2}+2p+2}\)

The inverse LaPlace of this is:

\(\displaystyle \L\\\fbox{e^{-t}(2cos(t)+3sin(t))}\)

I would say you just have a small algebra mistake somewhere. That's what it mostly is.
 
what i think i did wrong was assume that y(o)=2 is c1 and that y'(0)= 1 is c2... how do i determine the correct values for c1 and c2??
 
y(x) = e^(-x){Acos(x) + Bsin(x)}

y'(x) = -e^(-x){Acos(x) + Bsin(x)} + e^(-x){ - Asin(x) + Bcos(x)}......product rule
y'(0) = B - A

y(0) = A
 
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