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.