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I'm stuck on this problem! as usual.

Find a particular solution to

y'' + 4 y' + 4 y = 6.5 e^{-2 t}.

Here is my work:

y''+4y'+4y = 6.5e^(-2t)

r^2+4r+4 = 0;

r = -2, repeated root

So i made the guess:

y = A*t^2*e^(-2t);

y' = A(2t*e^(-2t) + t^2(-2)*e^(-2t))

y'' = A(2*e^(-2t)+t^2*4*e^(-2t))

So i equated co-efficents of e^(-2t) and got:

2A +4A = 6.5

6A = 6.5

A = 13/12

y = c1*e^(-2t)+c2*t*e^(-2t)+(13/12)*t^2*e^(-2t);

So as the answer i submitted was:

http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/42/a68981ceebd32318c1f96a3eee5ab31.png [Broken]

but it was wrong any ideas why?

Thanks!

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# Homework Help: Y''+4y'+4y = 6.5e^(-2t), I guessed y = At^2*e^(-2t) but still wrong! any ideas?

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