Euler Differential Equations

# Euler Differential Equations

Another type of second order differential equation that we can solve are known as Euler Differential Equations. What's particularly nice about Euler differential equations is that we can take a second order linear homogenous differential equation with non-constant coefficients and use a substitution to transform it into a second order linear homogenous differential equation with constant coefficients. We define Euler differential equations below.

 Definition: A second order differential equation in the form $t^2 \frac{d^2 y}{dt^2} + \alpha t \frac{dy}{dt} + \beta y = 0$, $t > 0$ where $\alpha, \beta \in \mathbb{R}$ is called an Euler Differential Equation.

In order to transform an Euler differential equation into a second order linear homogenous differential equation with constant coefficients, we will let $x = \ln t$. We will compute $\frac{dy}{dt}$ and $\frac{d^2y}{dt^2}$ as follows:

(1)
\begin{align} \quad \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt} = \frac{dy}{dx} \frac{1}{t} \end{align}
(2)
\begin{align} \quad \frac{d^2y}{dt^2} = \frac{d}{dt} \left ( \frac{dy}{dt} \right ) = \frac{d}{dt} \left ( \frac{dy}{dx} \frac{dx}{dt} \right ) = \frac{d}{dt} \left ( \frac{dy}{dx} \frac{1}{t} \right ) = \frac{d^2y}{dx^2} \frac{1}{t^2} - \frac{dy}{dx} \frac{1}{t^2} \end{align}

We will now plug this into our differential equation to get that:

(3)
\begin{align} \quad t^2 \left [ \frac{d^2y}{dx^2} \frac{1}{t^2} - \frac{dy}{dx} \frac{1}{t^2} \right ] + \alpha t \left [ \frac{dy}{dx} \frac{1}{t} \right ] + \beta y = 0 \\ \quad \frac{d^2 y}{dx^2} - \frac{dy}{dx} + \alpha \frac{dy}{dx} + \beta y = 0 \\ \quad \frac{d^2 y}{dx^2} + (\alpha - 1) \frac{dy}{dx} + \beta y = 0 \end{align}

Thus we have a second order linear homogenous differential equation with constant coefficients, namely, $1$, $\alpha - 1$, and $\beta$. We know how to solve differential equations of this type! Note that if $y = y_1(x)$ and $y = y_2(x)$ form a fundamental set of solutions to $\frac{d^2 y}{dx^2} + (\alpha - 1) \frac{dy}{dx} + \beta y = 0$ then $y = y_1(\ln t)$ and $y = y_2(\ln t)$ (which are obtained by reintroducing the substitution $x = \ln t$) form a fundamental set of solutions to the Euler equation $t^2 \frac{d^2 y}{dt^2} + \alpha t \frac{dy}{dt} + \beta y = 0$.

Let's now look at an example of solving an Euler differential equation.

## Example 1

Solve the Euler differential equation $t^2 \frac{d^2y}{dt^2} + 4t \frac{dy}{dt} + 2y = 0$ for $t > 0$.

Note that the second order linear homogenous differential equation above is indeed an Euler differential equation with $\alpha = 4$ and $\beta = 2$. Therefore we can immediately let $x = \ln t$ and rewrite this Euler differential equation as:

(4)
\begin{align} \quad \frac{d^2y}{dx^2} + (4 - 1) \frac{dy}{dx} + 2 y = 0 \\ \quad \frac{d^2 y}{dx^2} + 3 \frac{dy}{dx} + 2y = 0 \end{align}

The characteristic equation for this differential equation is $r^2 + 3r + 2 = 0$. This can easily be factored as $(r + 2)(r + 1) = 0$ and so we have two distinct real roots, $r_1 = -2$ and $r_2 = -1$. Therefore the general solution to this differential equation is:

(5)
\begin{align} \quad y = Ce^{-2x} + De^{-x} \end{align}

Making the substitution $x = \ln t$ back and we get a solution to the Euler differential equation:

(6)
\begin{align} \quad y = Ce^{-2\ln t} + De^{-\ln t} \\ \quad y = Ce^{\ln t^{-2}} + De^{\ln t^{-1}} \\ \quad y = Ct^{-2} + Dt^{-1} \\ \quad y = \frac{C}{t^2} + \frac{D}{t} \end{align}