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

In summary, the equation z4 = -1 has four roots, which are complex numbers with arguments of pi/4, 3pi/4, 5pi/4, and 7pi/4.
  • #1
Pr0x1mo
21
0
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?
 
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  • #2
The equation m4+1=0, does not have i as a root...
 
  • #3
Yes it does, which is the reason why wolfram alpha spat out the general solution using cos + sin (bernouilli).
 
  • #4
Maybe you should use i4+1 as input in wolfram alpha...
 
  • #5
lol ^ No, but thanks for your help.
 
  • #6
Pr0x1mo said:
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.
 
  • #7
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.
 
  • #8
but i is NOT a root of m4+1=0...
The roots of m4+1=0 are [tex]\pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i[/tex].

Try to put the following in wolfram alpha: x4+1=0. You'll see immediately what the roots are...
 
  • #9
micromass said:
but i is NOT a root of m4+1=0...
The roots of m4+1=0 are [tex]\pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i[/tex].

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 [tex]\pm\frac{1}{\sqrt{2}}\pm \frac{1}{\sqrt{2}}i[/tex] ??
 
  • #10
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
 
  • #11
Ok, but how did you solve the equation to get those roots?
 
  • #12
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.
 
  • #13
Awesome thanks
 

1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It is used to model change and is commonly used in physics, engineering, and other scientific fields.

2. What is the 4th derivative of y?

The 4th derivative of y is the derivative of the 3rd derivative of y, which is the derivative of the 2nd derivative of y, which is the derivative of the 1st derivative of y, which is the derivative of y itself.

3. How do you solve a differential equation?

There is no one method for solving all differential equations, as different types of equations require different techniques. However, some common methods include separation of variables, substitution, and using integrating factors.

4. What does the equation 4th derivative of y + y = 0 represent?

This equation represents a differential equation in which the 4th derivative of y plus the function y equals 0. It can be used to model various physical phenomena, such as the motion of a damped oscillator or the shape of a vibrating string.

5. What are some real-world applications of differential equations?

Differential equations are used to model a wide range of phenomena in science and engineering, including population growth, heat transfer, fluid dynamics, and electrical circuits. They are also used in fields such as economics, medicine, and ecology to analyze and predict complex systems.

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