Let y(n)(x) be the nth derivative of the unknown function y(x). Then a Cauchy–Euler equation of order n has the form
The substitution (that is, ; for , one might replace all instances of by , which extends the solution's domain to ) 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.[1]
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[1][2]
Observe that we can write the second-order Cauchy-Euler equation in terms of a linear differential operator as
where and is the identity operator.
We express the above operator as a polynomial in rather than . By the product rule,
So,
We can then use the quadratic formula to factor this operator into linear terms. More specifically, let denote the (possibly equal) values of
Then,
It can be seen that these factors commute, that is, . Hence, if , the solution to is a linear combination of the solutions to each of and , which can be solved by separation of variables.
Indeed, with , we have . So,
Thus the general solution is .
If , then we instead need to consider the solution of . Let , so that we can write
As before, the solution of is of the form . So, we are left to solve
We then rewrite the equation as
which one can recognize as being amenable to solution via an integrating factor.
Choose as our integrating factor. Multiplying our equation through by and recognizing the left-hand side as the derivative of a product, we then obtain
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 fm(n) as
Applying the difference operator to , we find that
If we do this k times, we find that
where the superscript (k) 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 N linearly independent solutions. Applying reduction of order in case of a multiple root m1 will yield expressions involving a discrete version of ln,
(Compare with: )
In cases where fractions become involved, one may use
instead (or simply use it in all cases), which coincides with the definition before for integer m.
^Boyce, William E.; DiPrima, Richard C. (2012). Rosatone, Laurie (ed.). Elementary Differential Equations and Boundary Value Problems (10th ed.). pp. 272–273. ISBN978-0-470-45831-0.