This Demonstration shows the solution to the Cauchy–Euler equation with initial conditions and and approximations to it using truncated series.

Assume solutions have the form

.

Take the first and second derivatives of this equation and substitute back into the original equation. If the equation is to be satisfied for all , the coefficient of each power of must be zero. This gives a quadratic equation in with roots and . Then the coefficients , , , … can be determined. The two solutions for are:

,

.

The final solution (plotted in blue) has the form , where and are determined by the initial conditions. With not too many terms it serves as a good approximation to the exact solution (plotted in red).

This example comes from Chapter 8 of [1] on series solutions and the Cauchy–Euler equation.

Reference

[1] J. R. Brannan and W. E. Boyce, Differential Equations with Boundary Value Problems: An Introduction to Modern Methods and Applications, New York: John Wiley and Sons, 2010.