# Euler equations

Here, we consider Euler equations: $$\tag{1} x^2 y’’ + a xy’ + by = 0, \quad x\ne 0,$$ where $a$, $b$ are constants.

## characteristic polynomial

Assuming that $y = x^r$ in (1) we obtain

$$\left[r(r-1) + a r + b\right] x^r = 0.$$

Since $x\ne 0$, $x^r\ne 0$, so the equation reduces to

$$\tag{2} r(r-1) + a r + b = 0.$$

(2) is called the characteristic polynomial of Euler equations. In principle, if (2) has a root $r_1$, then $y = x^{r_1}$ is a solution to (1).

But, Euler equations is a 2nd-order linear homogeneous ODE that has two linearly independent solutions. In the following we aim to find them all.

### A. Two distinct real roots

Auppose (2) has two roots $r = r_1$, $r_2$, the general solution of (1) can be written as

$$\tag{3} y(x) = c_1 x^{r_1} + c_2 x^{r_2},$$ where $c_1$, $c_2$ are constant.

#### Remark:

To calculate $x^{r}$, we usually use the following identity $$x^r = e^{r\ln x},$$ so that, even if $r\in Q^c$, $x^r$ can be computed easily.

### B. Two complex roots

Suppose (2) has two complex roots $r = \lambda \pm i\mu$, then $x^{\lambda \pm i\mu}$ is a solution. But what we want is real solutions, so we rewrite it as

\begin{align} x^{\lambda + i\mu} &= e^{(\lambda + i\mu)\ln x} \\ &= e^{\lambda \ln x}\left[\cos(\mu\ln x) + i\sin(\mu\ln x)\right] \\ &= x^r\cos(\mu\ln x) + ix^r\sin(\mu\ln x). \end{align}

We know that both the real and imaginary part are the solution to (1). Therefore, the general solution to (1) is

$$\tag{4} y(x) = c_1 x^{\lambda}\cos(\mu\ln x) + c_2 x^{\lambda}\sin(\mu\ln x),$$ where $c_1$, $c_2$ are constant.

### C. Only one root with multiplicity 2

Suppose there is only one root with multiplicity $2$, the characteristic polynomial must be in the form $(r-r_1)^2$, so that, by direct comparison with (2), we have $a = 1-2r_1$ and $b=r^2_1$. That is, Euler equation is given by

$$\tag{5} x^2 y’’ + (1 - 2r_1) xy’ + r^2_1 y = 0.$$

On the other hand, $y_1 = x^{r_1}$ must be a solution, and it is easy to check that $y_1$ satisfies the following 1st-order ODE:

$$\tag{6} \left(x\frac{d}{dx} - r_1\right) y = 0.$$

We have the following observations:

#### Observation 1

Use (6) and the fact that $r_1$ is a root with multiplicity $2$, one can easily see that (5) can be factored as

$$\tag{7} \left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y = 0.$$

#### Observation 2

Suppose we can find $y_2(x)$ such that

$$\tag{8} \left(x\frac{d}{dx} - r_1\right) y_2 = cy_1,$$ for some constant $c$, then we must have

$$\left(x\frac{d}{dx} - r_1\right)\left(x\frac{d}{dx} - r_1\right) y_2 = \left(x\frac{d}{dx} - r_1\right) cy_1 = 0.$$

#### Observation 3

Based on the idea of variation of parameters, since we have one solution being $x^{r_1}$, we can assume that $y_2 = x^{r_1}u(x)$ and pluging-in to (8). It turns out that $u(x)$ should satisfy

$$\tag{9} xu’ = cu,$$ and the solution is $u(x) = c\ln(x)$. Therefore, we have $y_2=x^{r_1}\ln(x)$.

Finally, the general solution is

$$\tag{10} y(x) = c_1 x^{r_1} + c_2 x^{r_1}\ln(x),$$ where $c_1$, $c_2$ are constant.

##### Te-Sheng Lin (林得勝)
###### Associate Professor

The focus of my research is concerned with the development of analytical and computational tools, and further to communicate with scientists from other disciplines to solve engineering problems in practice.