4th order differential equation

In summary, the general solution to the equation y''''-8y'=0 is y=c1+c2*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))], where the roots are r=0,2,-1+i*sqrt(3),-1-i*sqrt(3). This is derived from the characteristic polynomial r^4-8r=0, where the roots are r=0,2,-1+i*sqrt(3),-1-i*sqrt(3) over the field of complex numbers. The root e(2*m*pi*i)/3 is derived from Euler's formula and represents
  • #1
chuy52506
77
0
I'm trying to find the gen. solution to the equation y''''-8y'=0
I found the characteristic polynomial by plugging in ert as a solution to y.
I got,
r^4-8r=0
I simplified to get
r*(r^3-8)
Thus one root is 0, for the other 3 i must find the cubed root of 8.
I know the answer is 2*e2m*pi*i/3 for m=0,1,2
How do I arrive at that answer?
I tried the following:
Represent 8 as 8=8[cos(2*pi)+i*sin(2*pi)]=8*ei*pi
 
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  • #2
chuy52506 said:
I'm trying to find the gen. solution to the equation y''''-8y'=0
I found the characteristic polynomial by plugging in ert as a solution to y.
I got,
r^4-8r=0
I simplified to get
r*(r^3-8)
Thus one root is 0, for the other 3 i must find the cubed root of 8.
I know the answer is 2*e2m*pi*i/3 for m=0,1,2
How do I arrive at that answer?
I tried the following:
Represent 8 as 8=8[cos(2*pi)+i*sin(2*pi)]=8*ei*pi
It should be 8 = 8ei*2πm, for any m, right?
 
  • #3
The roots are r=0,2*e(2*m*pi*i)/3
then it says this is equivalent to
r=0,2,-1+i*sqrt(3),-1-i*sqrt(3)

Then the gen solution is
y=c1+c2*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))]

I don't know how they arrive to this
 
  • #4
Obviously, the cubic root of 8 is 2.
r^4-8r = r(r-2)(r²+2r+4)
= r (r-2) [r+1+i sqrt(3)] [r+1-i sqrt(3)]
 
  • #5
chuy52506 said:
The roots are r=0,2*e(2*m*pi*i)/3
then it says this is equivalent to
r=0,2,-1+i*sqrt(3),-1-i*sqrt(3)

Then the gen solution is
y=c1+c2*e2*t+e-t*[c3*cos(t*sqrt(3))+c4*sin(t*sqrt(3))]

I don't know how they arrive to this
Suppose λ is a root of the polynomial. That is, the differential equation has a factor (D-λ), where D = d/dt, so the equation can be written P(D)(D-λ)y = 0, for some polynomial P.
Try the solution y = eλt: P(D)(D-λ)eλt = P(D)(D(eλt - λeλt) = P(D)(λeλt - λeλt) = 0. So the general solution is a linear combination of such terms.
A complication arises when there is a repeated root, i.e. a factor (D-λ)n. It's fairly easy to show that treλt is also a solution for r = 1, .. n-1.
 
  • #6
But where is the root e(2*m*pi*i)/3 derived from? and how is this equivalent to the roots:2,-1+i*sqrt(3),-1-i*sqrt(3) for m=1,2,3
 
  • #7
exp(2∏n i)=1
so
(exp(2∏n i/3))^3=1
also
exp(2∏n ix)=cos(2∏n x)+i sin(2∏n x)
so
exp(2∏n i/3)=cos(2∏n /3)+i sin(2∏n /3)
 
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  • #8
lurflurf said:
exp(2∏n i)=1
so
(exp(2∏n i/3))^3=1
also
exp(2∏n ix)=cos(2∏n ix)+i sin(2∏n ix)
so
exp(2∏n i/3)=cos(2∏n i/3)+i sin(2∏n i/3)
Small correction (too many i's):
exp(2∏n ix)=cos(2∏n x) + i sin(2∏n x)
so
exp(2∏n i/3)=cos(2∏n/3) + i sin(2∏n/3)
... and cos(2∏/3) = -cos(∏/3) = -(√3)/2 etc.
 
  • #9
chuy52506 said:
But where is the root e(2*m*pi*i)/3 derived from? and how is this equivalent to the roots:2,-1+i*sqrt(3),-1-i*sqrt(3) for m=1,2,3

What you are getting confused about is the field of the reals vs. the field of complex numbers. Over the field of the reals there is only one solution to the equation r^3-8=0 namely r=2, but over the field of complex numbers there are 3 distinct solutions to the equation r^3-8=0.

To see this understand that [tex]exp(i2m\pi)=1[/tex] for any [tex]m\epsilon\mathbb{Z}[/tex]. So if, [tex]r^3=8\cdot1=8exp(i2m\pi)[/tex] then
[tex]r =2exp(\frac{i2m\pi}{3})[/tex] but you can see the only distinct ones are for m= 0,1,2 since for m beyond or below that you repeat your answers. To answer your other question you need to know about Euler's formula which says that: [tex]e^{i\phi}=cos(\phi)+isin(\phi)[/tex]

So for example, [tex]2exp(i\frac{2\pi}{3})=2(cos(\frac{2\pi}{3})+i\sin({\frac{2\pi}{3}})=2(-\frac{1}{2}+i\frac{\sqrt{3}}{2})[/tex]
 
Last edited:

1. What is a 4th order differential equation?

A 4th order differential equation is a mathematical equation that involves a 4th derivative of an unknown function. It can be written in the form of y'''' = f(x, y, y', y'', y'''). This type of equation is commonly used in physics and engineering to model systems with multiple variables and complex behaviors.

2. How is a 4th order differential equation solved?

Solving a 4th order differential equation involves finding a function that satisfies the equation. This can be done analytically by using various methods such as separation of variables, variation of parameters, or the method of undetermined coefficients. It can also be solved numerically using techniques like Euler's method or Runge-Kutta methods.

3. What are the applications of 4th order differential equations?

4th order differential equations have numerous applications in various fields of science and engineering. They are used to model physical phenomena such as motion, heat transfer, and electrical circuits. They are also used in fields like economics, biology, and chemistry to study complex systems and their behaviors.

4. Can a 4th order differential equation have multiple solutions?

Yes, a 4th order differential equation can have multiple solutions. This is because it is a higher-order equation with more variables, making it more complex and allowing for a wider range of possible solutions. These solutions can also be affected by initial conditions and boundary conditions.

5. Are there any real-world examples of 4th order differential equations?

Yes, there are many real-world examples of 4th order differential equations. For example, the equation for a damped harmonic oscillator in physics is a 4th order differential equation. In engineering, the equation for the deflection of a beam under a load is also a 4th order differential equation. These equations can be used to model and analyze real-world systems and phenomena.

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