Differential equation question, 4th derivative of y + y = 0

[Pr0x1mo]

Ok, the question/equation is

y^4 + y = 0

Now i know that the characteristic equation is m^4 + 1 = 0

and if i would to solve that it would give me plus or minus 1i as a complex root. so the general solution would be:

c1cos(x) + c2sin(x) but i know this is not the answer because since its a 4th degree, it should have 4 roots, repeated or not.

When i rant it thru wolframalpha, it gave me this: http://www.wolframalpha.com/input/?i=d^4y%2Fdx^4+%2B+y+%3D+0

How did it get to that answer?

 

[micromass]

The equation m⁴+1=0, does not have i as a root...

 

[Pr0x1mo]

Yes it does, which is the reason why wolfram alpha spat out the general solution using cos + sin (bernouilli).

 

[micromass]

Maybe you should use i⁴+1 as input in wolfram alpha...

 

[Pr0x1mo]

lol ^ No, but thanks for your help.

 

[Mark44]

Yes it does, which is the reason why wolfram alpha spat out the general solution using cos + sin (bernouilli).

No it doesn't. i is not a root of m^4 + 1 = 0. i^4 = (-1)^2 = +1. Replacing m by i in the equation m^4 + 1 = 0 gives you 1 + 1 != 0.

The four roots are complex numbers whose arguments (angles) are pi/4, 3pi/4, 5pi/4, 7pi/4.

 

[Pr0x1mo]

The reason i know that there are i's in the solution is because this question came from the chapter in my book titled "Characteristic equations and complex roots," did you take a look at the solution that wolframalpha displayed? I know that this is the answer because my Professor alluded to hints that there would be more than one e^x(c1cosx + c2sinx) because of repeated roots.

I just want to figure out how the equation was rewritten to get plus or minus 1/(2)^1/2 i as a root.

 

[micromass]

but i is NOT a root of m⁴+1=0...
The roots of m⁴+1=0 are \pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i.

Try to put the following in wolfram alpha: x⁴+1=0. You'll see immediately what the roots are...

 

[Pr0x1mo]

but i is NOT a root of m⁴+1=0...
The roots of m⁴+1=0 are \pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i.

Try to put the following in wolfram alpha: x⁴+1=0. You'll see immediately what the roots are...

Ok, see, that was my entire question how did they get plus or minus \pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i ??

 

[Mark44]

micromass and I are pretty much saying the same thing, but in different forms. He has shown the roots in Cartesian form (a + bi), and I said what they were in a sort of polar form.

One of the roots is cos(pi/4) + i*sin(pi/4), or 1/sqrt(2) + 1/sqrt(2) * i.
The other roots are
cos(3pi/4) + i*sin(3pi/4) = -1/sqrt(2) + 1/sqrt(2) * i
cos(5pi/4) + i*sin(5pi/4) = -1/sqrt(2) - 1/sqrt(2) * i
cos(7pi/4) + i*sin(7pi/4) = 1/sqrt(2) - 1/sqrt(2) * i

 

[Pr0x1mo]

Ok, but how did you solve the equation to get those roots?

 

[Mark44]

First, by changing -1 to its polar form of cos(pi) + i sin(pi), which is really cos(pi + 2k *pi) + i sin(pi + 2k*pi), and
second, by using the Theorem of de Moivre, which says that [r(cos(x) + i sin(x)]⁴ = rⁿ(cos(n *x) + i sin(n * x))

Since z⁴ = -1, then z = (-1)^1/4,
so z = [cos(pi + 2k *pi) + i sin(pi + 2k*pi)]^1/4
= cos(pi/4 + 2k/4 *pi) + i sin(pi/4 + 2k/4*pi)
= cos(pi/4 + k * pi/2) + i sin(pi/4 + k * pi/2)

The formula above holds for all integer values of k, but there are only four distinct roots of the equation z⁴ = -1, corresponding to k = 0, 1, 2, and 3. When k >= 4, the roots start repeating.

 

[Pr0x1mo]

Awesome thanks