# Euler substitution

The Euler substitution is a useful substitution for solving linear homogeneous ordinary differential equations with constant coefficients.[1] These are differential equations of the form:



The solution will be a linear combination of n linearly independent eigenfunctions, :



The Euler substitution allows one to determine these eigenfunctions. It simplifies the problem as instead of having to solve a differential equation, one must instead solve a polynomial

## Method

Consider solving a general equation of the form above:



The Euler substitution is . This yields, for the first few derivatives:



Substituting back in, this yields:



Dividing through by ,



This is a polynomial and very easy to solve. If it has solutions λi, the solution to the differential equation can be written as:



Where the ci are arbitrary constants. If a root is repeated, then there will not be n linearly independent solutions. In this case, if the root is repeated m times, then m linearly independent functions can be created by multiplying  by 1, t, t2...tm-1.

In the important case of n=2, a second order equation, this polynomial can be written as . This equation can easily be solved using the quadratic formula. There are three cases that arise:

Case I: When  In this case, the solutions are exponentials:



Case II: When  In this case the solutions are complex exponentials. These can be rewritten in terms of sines and cosines using Euler's formula:

 and ,


When Case III:  In this case there is only a single solution for λ as it is a repeated root. Therefore  is the same as  and so are not linearly independent. Using the above, m=2 and so we can multiply by 1 and t. Therefore, for this case the general solution is:



A particular solution can then be found by applying boundary conditions.

## Example

Consider the differential equation:



Substituting in  in and then dividing by  produces:



This only has one one solution, namely λ=1. This is a repeated root, so we multiply one solution by t to form 2 linearly independent solutions. Thus the general solution is:



## References

1. K.F. Riley, M.P. Hobson, S.J. Bence, Mathematical Methods for Physics and Engineering, Cambridge University Press, 3rd ed., 2006