# 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

## Contents

## 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 λ_{i}, the solution to the differential equation can be written as:

Where the c_{i} 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, t^{2}...t^{m-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

- ↑ K.F. Riley, M.P. Hobson, S.J. Bence,
*Mathematical Methods for Physics and Engineering*, Cambridge University Press, 3^{rd}ed., 2006