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).