Ode Again!

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!

 

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?

 

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.)