# Properties of ordinary differential equations - solution

I know that we take characteristic polynomial of forms like $\lambda^2 - \lambda - 2$ to find out the solution of ordinary differential equation of the form $e^{\lambda x}$ - conjugate ones can be slightly modified. The question is, there are other forms of solution, right? So, why do people just use superposition of natural exponentiation forms when other solutions are available? Or am I mistaken?

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There are no more. –  André Nicolas Dec 25 '12 at 0:13

Since we are talking about homogeneous systems of ﬁrst order diﬀerential equations, the solutions always use exponentials, but you can have several variants of eigenvalues that modify how you write the linear combination of solutions.

When you have real, distinct eigenvalues, the solution is a linear combination of exponentials using the eigenvalues and eigenvectors. See these notes for examples of this type.

When you have complex eignevalues, you can write the solution as a linear combination of complex eigenvalues, or using Euler's formula, as $e^{(a + ib)t}$ = $e^{at}cos(bt)$ + $ie^{(at)}sin(bt)$. See these notes for examples of this type.

When you have eigenvalues with varying geometric and algebraic multiplicity (sometimes called defective eigenvalues), that is, repeated eigenvalues, things get more interesting, so you want to look up what those terms mean. See these notes for examples of this type.

However, like Andre replied, they are always a form of the exponential because of the type of system we are dealing with. However, when you get to nonlinear systems, things can change drastically.

Regards

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It also changes for systems with nonconstant coefficients. –  Robert Israel Dec 25 '12 at 0:51
@Robert Israel - thank you for adding that note! –  Amzoti Dec 25 '12 at 0:53
What about ones that are not first-order? –  EFR Dec 25 '12 at 1:23
Higher-order systems are equivalent to first-order systems of higher dimension. You just have to add more variables to stand for derivatives of the original variables. –  Robert Israel Dec 25 '12 at 1:31
$\;+1^+\;\;\ddot\smile \;\;\diamondsuit$ –  amWhy May 11 '13 at 0:17