# Euler Differential Equations Examples 1

Recall from the Euler Differential Equations page that a second order differential equation in the form $t^2 \frac{d^2y}{dt^2} + \alpha t \frac{dy}{dt} + \beta y = 0$, $t > 0$, and $\alpha, \beta \in \mathbb{R}$ is called an Euler differential equation.

If we set $x = \ln t$ then we can transform the differential equation above to:

(1)These types of second order differential equations can be solved easily with the techniques seen from earlier. Let's now look at some more examples of Euler differential equations.

## Example 1

**Solve the Euler differential equation $t^2 \frac{d^2y}{dt^2} - 3t \frac{dy}{dt} + 4y = 0$ for $t > 0$.**

In this case we have that $\alpha = -3$ and $\beta = 4$. Let $x = \ln t$. Then we can transform the differential equation above into:

(2)The characteristic equation for this differential equation is:

(3)Therefore the characteristic equation have one real distinct root, $r_1 = 2$. Therefore the general solution to the modified differential equation for $C$ and $D$ as constants is:

(4)We now reintroduce the substitution $x = \ln t$ to get:

(5)## Example 2

**Solve the Euler differential equation $t^4 \frac{d^2y}{dt^2} + 5t^3 \frac{dy}{dt} + 13t^2y = 0$ for $t > 0$.**

The differential equation above may not look like an Euler differential equation, but if we divide both sides by $t^2$ (noting that $t \neq 0$ since $t > 0$ then we get the equivalent differential equation:

(6)This differential is clearly an Euler differential equation with $\alpha = 5$ and $\beta = 13$. Let $x = \ln t$. Then we have that an equivalent differential equation with this substitution is given by:

(7)The characteristic equation for this differential equation is $r^2 + 4r + 13 = 0$. We can use the quadratic formula to find the roots of this characteristic equation. We have that:

(8)Therefore the characteristic equation has two complex roots, $r_1 = -2 + 3i$ and $r_2 = -2 - 3i$. The general solution to the modified differential equation for $C$ and $D$ as constants is therefore:

(9)Using the substitution that $x = \ln t$ and we have that:

(10)