Let y be the nth derivative of the unknown function y. Then a Cauchy–Euler equation of order n has the form The substitution may be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution may be used to directly solve for the basic solutions.
Second order – solving through trial solution
The most common Cauchy–Euler equation is the second-order equation, appearing in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. The second order Cauchy–Euler equation is We assume a trial solution Differentiating gives and Substituting into the original equation leads to requiring Rearranging and factoring gives the indicial equation We then solve for m. There are three particular cases of interest:
In case #1, the solution is In case #2, the solution is To get to this solution, the method of reduction of order must be applied after having found one solution y = xm. In case #3, the solution is For ∈ ℝ. This form of the solution is derived by setting x = et and using Euler's formula
We operate the variable substitution defined by Differentiating gives Substituting the differential equation becomes This equation in is solved via its characteristic polynomial Now let and denote the two roots of this polynomial. We analyze the two main cases: distinct roots and double roots: If the roots are distinct, the general solution is If the roots are equal, the general solution is In both cases, the solution may be found by setting. Hence, in the first case, and in the second case,
Example
Given we substitute the simple solution xm: For xm to be a solution, either x = 0, which gives the trivial solution, or the coefficient of xm is zero. Solving the quadratic equation, we get m = 1, 3. The general solution is therefore
There is a difference equation analogue to the Cauchy-Euler equation. For a fixed m > 0, define the sequence ƒm as Applying the difference operator to, we find that If we do this k times, we find that where the superscript denotes applying the difference operator k times. Comparing this to the fact that the k-th derivative of xm equals suggests that we can solve the N-th order difference equation in a similar manner to the differential equation case. Indeed, substituting the trial solution brings us to the same situation as the differential equation case, One may now proceed as in the differential equation case, since the general solution of an N-th order linear difference equation is also the linear combination of Nlinearly independent solutions. Applying reduction of order in case of a multiple rootm1 will yield expressions involving a discrete version of ln, In cases where fractions become involved, one may use instead, which coincides with the definition before for integer m.
