what is y?

$\displaystyle y^(5)+2y=0$

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- Oct 26th 2009, 05:39 AM #1

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- Oct 26th 2009, 06:00 AM #2

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- Oct 26th 2009, 06:38 AM #3

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- Oct 26th 2009, 07:40 AM #4

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Use a trial solution $\displaystyle e^{\lambda t}$ , this gives you a charateristic equation:

$\displaystyle \lambda^5+2=0$

so $\displaystyle \lambda=(-2)^{1/5}$ that is the general solution is:

$\displaystyle y(t)=Ae^{\lambda_1t}+Be^{\lambda_2t}+Ce^{\lambda_3 t}+De^{\lambda_4t}+Ee^{\lambda_5t}$

where the $\displaystyle \lambda_i $'s are the fith roots of $\displaystyle -2$

CB