# Euler equations

# 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.