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I'm teaching basic differential equations this semester and I was wondering if there were methods of solving systems of linear differential equations that don't use (directly) eigenvectors/eigenvalues. Is there such a way?

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    You could turn them all into 1D higher order systems and use the characteristic polynomial. – GPerez May 15 '15 at 19:16
  • Use a solution $t \mapsto e^{\lambda t}$ and solve for $\lambda$? Not sure how to handle repeated roots. – copper.hat May 15 '15 at 19:20
  • @GPerez Yes, of course. I was somehow thinking of keeping the problem as a system of equations without doing this. My question is not defined well. –  May 15 '15 at 19:22
  • @copper.hat Can you elaborate a bit on this? I'm not seeing it right away. –  May 15 '15 at 19:23
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    (I haven't thought this through at all, I just threw out something that popped into my head!) If you have a differential equation $\sum_k \alpha_k D^k x = 0$, then since $D^k (t \mapsto e^{\lambda t}) = t \mapsto \lambda^k e^{\lambda t}$, then if $t \mapsto e^{\lambda t}$ is a solution, the $\lambda$ must satisfy $\sum_k \alpha_k \lambda^k = 0$. This introduces the characteristic polynomial. – copper.hat May 15 '15 at 19:26
  • Just spitballing here, but maybe calculate the characteristic polynomial of the matrix $A$ of coefficients of the ODE, and use Cayley-Hamilton to calculate successive powers of $A$. In theory you could then calculate $e^{At}$ directly. Sounds tough though. – GPerez May 15 '15 at 19:44
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    i don't know if characteristic polynomial can be avoided; the putzers algorithm of finding the exponential matrix avoids finding the eigenvectors. – abel May 15 '15 at 19:45
  • @abel Thanks for the reference to putzers algorithm. I was not aware of that. –  May 15 '15 at 20:55
  • here is my attempt at explaining putters method:http://math.stackexchange.com/questions/33851/how-to-calculate-the-matrix-exponential-explicitly-for-a-matrix-which-isnt-diag/35139#35139 – abel May 15 '15 at 21:33

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I know my question was a bit malformed. What I'm thinking is how would one approach solving a system of linear homogeneous differential equations without knowledge of linear algebra.

I was experimenting with something along the lines of looking at lines through the origin as solutions and from these constructing the general solution. Lines through the origin can be characterized as

$x_{i}(t) = m_{i} x_{1}(t)$

for $i$ from 2 to $n$. Now use that

$\frac{dx_{i}}{dx_{1}} = m_{i}$

and that

$\frac{dx_{i}}{dt} = \left(\sum a_{ij} m_{i}\right)x_{1}(t)$

with $m_{1}=1$ to get a system of equations in the $m_{i}$. Note that $x_{1}(t)$ falls out. Once one finds the solutions to the $m_{i}$ we can find a solution to the system of equations. This will lead to finding the eigenvalues but without use of them explicitly.

An example. Consider

A = {{1, -1}, {1, 1}} (in Mathematica notation).

$y(t) = m x(t)$

$\frac{dx}{dt} = x(t) - m x(t)$

$\frac{dy}{dt} = x(t) + m x(t)$

Thus

$\frac{dy}{dx} = m = \frac{x(t) +m x(t)}{x(t) - m x(t)}=\frac{1+m}{1-m}$

Solutions to $m = \frac{1+m}{1-m}$ are $-i$ and $i$. This gives us the equation

$\frac{dx}{dt} = x(t)+i x(t)$

Which has a solution of $e^{(1+i)t}$ and $1+i$ is an eigenvalue of $A$.

I'm finding eigenvalues of $A$ by looking at solutions to the system of differential equation that go through the origin. I run into trouble with eigenvalues of multiplicity greater than 1. I was wondering if there was a general technique along these lines that is more formal that what I'm doing.

  • the only solutions that have line trajectories are along the egenvectors. – abel May 15 '15 at 21:34
  • Yes. Thus I'm using a system of differential equations to find eigenvalues and eigenvectors rather than the reverse. That's what got me started along this line of reasoning. –  May 15 '15 at 22:27