Need help with 4th order differential equation

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nowak
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Homework Statement



y^(4)+y=0

Homework Equations



Need to use de Moivre's formula to obtain answer.

3. The Attempt at a Solution .

I can obtain the characteristic equation r^4=-1. From there I tried saying y_1 = cosx and y_2=sinx. However, my book has the answer listed as y= [e^(√(2)x/2)(cos(√(2)/2)x+sin(√(2)/2)x)]+[e^-(√(2)x/2)(cos(√(2)/2)x+sin(√(2)/2)x)]. I am unsure how they obtained this answer. When I asked my teacher he said I had to use the de Moivre formula, but I am unsure how to apply this. PLEASE HELP!
 
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I said r^2(r^2)=0 so I took the first root to be (0+i) and the second to be (0-i), from there I can say y=e^(0)(Ccosx+C'sinx). I was unsure how to get the other two roots from there. Tried reduction of order saying that if y_1=cosx then v_1y_1=y_3 but this does not give the answer in the book.
 
I know, I only knew how to solve for the two of them. I tried to find the other two using the reduction of order method and was unable to get the answer the book said.
 
nowak said:

Homework Statement



y^(4)+y=0

Homework Equations



Need to use de Moivre's formula to obtain answer.

That's what you need to do then. You have the equation, and I'll use z:

[tex]z^4=-1[/tex]

or:

[tex]z=\sqrt[4]{-1}[/tex]

and de Moivre's formula allows you to solve for the four roots using the expression:

[tex]z=|r|^{1/4}e^{i/4(\theta+2n\pi)}[/tex]

but I think though you may not know how to apply it. r is the absolute value of -1 or just 1 right. Theta is the argument of -1. Well, that's just pi. And n goes from 0 to 3.