# General solutionsof linear second order homogenous diff equations

• Feb 14th 2008, 09:24 AM
moolimanj
General solutionsof linear second order homogenous diff equations
Hi all

• Feb 14th 2008, 10:11 AM
topsquark
Quote:

Originally Posted by moolimanj
Hi all

Let me do the first to demonstrate the method. Then you can try the others.

$\displaystyle 5 \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} + y = 0$

You first need to solve the characteristic equation. This is the polynomial equation (for m, according to the book I was taught from, but you may naturally use whatever variable you like) given by the coefficients of the differential equation. Powers of m are given by the order of the term.
$\displaystyle 5m^2 + 4m + 1 = 0$

Solving this equation gives:
$\displaystyle m = \frac{-2}{5} \pm \frac{i}{5}$

Now, each possible solution for m is going to give a linearly independent solution for the homogeneous equation of the form: $\displaystyle e^{m_n \cdot x}$ where the $\displaystyle m_n$ is the "nth" possible value of m given by the characteristic equation. So in this case the most solution for the homogeneous differential equation will be:
$\displaystyle y(x) = A \cdot exp \left [ \left ( \frac{-2}{5} + \frac{i}{5} \right ) x \right ] + B \cdot exp \left [ \left ( \frac{-2}{5} - \frac{i}{5} \right ) x \right ]$

There are other ways to write this, using sine and cosine to replace the complex exponentials, but I'll leave it this way as it exposes the structure behind the method.

(Note: the other two problems solve much more nicely than this one.)

-Dan