Euler substitution

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

<math>

a_n\frac{d^n y}{dx^n} + a_{n-1}\frac{d^{n-1} y}{dx^{n-1}} +...+ a_1\frac{d y}{dx} + a_0 y =0 </math>

The solution will be a linear combination of n linearly independent eigenfunctions, <math>y_1(t),y_2(t),...y_n(t)</math>:

<math>

y(t) = c_1y_1(t) + c_2y_2(t) +...+ c_ny_n(t) </math>

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:

<math>

a_n\frac{d^n y}{dx^n} + a_{n-1}\frac{d^{n-1} y}{dx^{n-1}} +...+ a_1\frac{d y}{dx} + a_0 y =0 </math>

The Euler substitution is <math>y = e^{\lambda t}</math>. This yields, for the first few derivatives:

<math>y = e^{\lambda t}, \, \frac{d y}{dx} = \lambda e^{\lambda t}, \, \frac{d^2 y}{dx^2} = \lambda^2 e^{\lambda t}</math>

Substituting back in, this yields:

<math>a_n\lambda^n e^{\lambda t} +...+ a_{n-1}\lambda e^{\lambda t} + a_0e^{\lambda t} = 0</math>

Dividing through by <math>e^{\lambda t}</math>,

<math>a_n +...+ a_{n-1} + a_0 = 0</math>

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:

<math>

y(t) = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} +...+ c_n e^{\lambda_n t} </math>

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 <math>e^{\lambda t}</math> by 1, t, t²...t^m-1.

In the important case of n=2, a second order equation, this polynomial can be written as <math>a\lambda^2+b\lambda+c=0</math>. This equation can easily be solved using the quadratic formula. There are three cases that arise:

Case I: When <math>\sqrt {b^2-4ac} > 0</math> In this case, the solutions are exponentials:

<math>y = c_1 y_1 + c_2 y_2 = c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t} </math>

Case II: When <math>\sqrt {b^2-4ac} < 0</math> In this case the solutions are complex exponentials. These can be rewritten in terms of sines and cosines using Euler's formula:

<math>\lambda_1 = r + i \mu</math> and <math>\lambda_2 = r - i \mu</math>,

<math>y = c_1 y_1 + c_2 y_2 = e^{rt}(c_1 cos{\mu t} + c_2 sin{\mu t})</math>

When Case III: <math>\sqrt {b^2-4ac} = 0</math> In this case there is only a single solution for λ as it is a repeated root. Therefore <math>y_1(t)</math> is the same as <math>y_2(t)</math> 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:

<math>y = c_1 y_1 + c_2 y_2 = e^{\lambda t}(c_1 + c_2 t)</math>

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

Example

Consider the differential equation:

<math>

y-2y'+y=0 </math>

Substituting in <math>y=e^{\lambda t}</math> in and then dividing by <math>y=e^{\lambda t}</math> produces:

<math>

\lambda^2 - 2\lambda + 1=0 </math>

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:

<math>

y(t) = c_1 e^{t} + c_2 te^t </math>

See also

• Exact differential equation