Euler Differential Equations Examples 1

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

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

The characteristic equation for this differential equation is:

(3)
\begin{align} \quad r^2 - 4r + 4 = 0 \\ \quad (r - 2)(r - 2) = 0 \end{align}

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)
\begin{align} \quad y = Ce^{2x} + Dxe^{2x} \end{align}

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

(5)
\begin{align} \quad y = Ce^{2\ln t} + Dte^{2\ln t} \\ \quad y = Ce^{\ln t^2} + D\ln (t) e^{\ln t^2} \\ \quad y = Ct^2 + D \ln (t) t^2 \end{align}

## 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)
\begin{align} \quad t^2 \frac{d^2 y}{dt^2} + 5t \frac{dy}{dt} + 13y = 0 \end{align}

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)
\begin{align} \quad \frac{d^2 y}{dx^2} + (\alpha - 1) \frac{dy}{dx} + \beta y = 0 \\ \quad \frac{d^2 y}{dx^2} + 4 \frac{dy}{dx} + 13 y = 0 \end{align}

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)
\begin{align} \quad r = \frac{-4 \pm \sqrt{16 - 4(1)(13)}}{2} = \frac{-4 \pm \sqrt{-36}}{2} = \frac{-4 \pm 6i}{2} = -2 \pm 3i \end{align}

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)
\begin{align} \quad y = Ce^{-2x} \cos (3x) + De^{-2x} \sin (3x) \end{align}

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

(10)
\begin{align} \quad y = Ce^{-2 \ln t} \cos (3 \ln t) + De^{-2 \ln t} \sin (3 \ln t) \\ \quad y = Ce^{\ln t^{-2}} \cos (\ln t^3) + De^{\ln t^{-2}} \sin (\ln t^3) \\ \quad y = \frac{C}{t^2} \cos (\ln t^3) + \frac{D}{t^2} \sin (\ln t^3) \end{align}