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Find the general solution of $y^{(6)}+2y^{(4)}+y'' = 0$.

$r^6+2r^4+r^2=0$

$r^2(r^4+2r^2+1)=0$

$r^2[(r^2+1)(r^2+1)]$

So we have the roots:

$0$: Multiplicity 2

$+i$: Multiplicity 2

$-i$: Multiplicity 2

Now I'm not sure. I'm supposed to arrive at:

$y(x) = c_1+c_2x+c_3\cos x+c_4\sin x+c_5x\cos x+c_6x\sin x$

EDIT: I'm particularly curious as to what the multiplicity does to the general solution.

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    It's the multiplicity that necessitates the factor of $x$ in the 2nd, 5th, and 6th terms in the answer.2012-10-06
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    I'm confused by the "sixth degree" in the title and the $y^6$; this suggests a horrendous non-linear differential equation. Or should I read "sixth order" and $y^{(6)}$ (as well as $y^{(4)}$)?2012-10-06
  • 0
    Corrected to sixth order.2012-10-06

3 Answers 3