# 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)

(2)

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

(3)

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)

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)

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

(6)

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