Ode Again!

  • Thread starter heman
  • Start date
  • #1
349
0

Main Question or Discussion Point

Here is another problem which is driving me insane!I just need a hint on how to proceed!

To Solve:-
p^7 + p^3 - p^2 + 1 =0

here p = dy/dx
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Since that is a linear homogeneous d. e., it characteristic equation is just what you give: p7+ p3- p2+ 1= 0. There are no general ways of solving 7th degree polynomial equations so the best you could hope for is some simple equation. Because the leading and ending coefficients are both 1, the only possible rational solutions are p= 1 and p= -1- and neither of those satisfy the equation!

My suggestion- go insane!
 
  • #3
349
0
Halls,You triggered in the right direction!
I asked my tutor and he said that there will be atleast 1 real root
=> p=k
or y=kx + c
From here i am thinking ahead!
 
  • #4
saltydog
Science Advisor
Homework Helper
1,582
3
I have a question:

So we look for solutions of the form y=mx+b, plug it in, get a polynomial in m and solve for the roots. So there's one real root. So we find it numerically (I ain't proud), and then we get the solution:

[tex]y(x)=m_rx+b[/tex]

So, how do we know if there are other solutions?
 
  • #5
349
0
saltydog said:
So, how do we know if there are other solutions?
The same thing i asked him!
He said our aim is to get just 1 family of curves of solution!
The other six can be complex,real who knows!
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
955
heman, I interpreted "p^7" where p= dy/dx to mean the seventh derivative (operator notation). saltydog tells me that he interpreted it as the seventh power of dy/dx so that this is not a linear equation at all. Which is it?

(Yes, the polynomial p^7 + p^3 - p^2 + 1 =0, by DesCartes' "rule of signs", has one negative real root and either 0 or 2 positive real roots. That is, the equation may have (1) 1 negative real root and 3 pair of conjugate complex roots or (2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots. However, the real roots are not rational.)
 
  • #7
349
0
Ohhhh....I am the source of miscommunication. :surprised ...Halls,it's 7th power,i should have had clarified more to avoid this..! o:)

Your Predictions!!
(1) 1 negative real root and 3 pair of conjugate complex roots
I understand that it can have 3 pair of conjugate complex roots-but how do you say that real root will be negative?

(2) 1 negative real root, 2 positive real roots, and 2 pair of conjugate comples roots
How are you able to tell the signs??

I have taken the Complex Analysis Course last semester but i haven't come across any such theorem which tells about the sign!
 

Related Threads on Ode Again!

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
722
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
5
Views
6K
Top