$$
\color{red}{x^2}\frac{\color{red}{d^2}y}{dx^2} - 2\color{red}{x}\frac{\color{red}{d}y}{dx} - 4y = 0\\
4\color{red}{x^2}y\color{red}{''} + 8\color{red}{x}y\color{red}{'} + y = 0\\
$$

are examples of Cauchy-Euler differential equations. A linear differential equation of the form

$$ a_n x^n \frac{d^n y}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \dots + a_1 x \frac{dy}{dx} + a_0y = g(x), $$where the coefficients $a_n, a_{n-1}, \dots, a_0$ are constants, is known as a Cauchy-Euler equation. The important observation is that coefficient $x^k$ matches the order of differentiation.

**Cauchy-Euler differential equations have solutions in format
$\boldsymbol{y(x) = x^m}$** and are easily solved. The solution is quite straight-forward. We
lay expected solution $y(x) = x^m$ and feed the solution with its derivatives
$y'(x)$ and $y''(x)$ into DE. Solving leads to separating $x^m$ and **auxiliary
equation (see the examples below).** The roots $m$ have to be determined from auxiliary equation.

Once we have roots $m$, we have to deal with 3 possibilities:

Solution is in the form $y(x) = c_1 x^{m_1} + c_2 x^{m_2}$.

E.g. $m_1 = 2$, $m_2 = 2$. The solutions can not be
linearly dependent. *The Reduction of order method* determines from the first solution $y_1(x) = x^{m}$
the second solution as $y_2(x) = x^{m} \log x$.

It is an analogy to the above. The imaginary roots come in conjugate pairs and solution can be written as $y(x) = c_1 x^{\alpha - i \beta} + c_2 x^{\alpha + i \beta}$. However we often prefer to work with expression without complex numbers. The other form is $y = C_1 x^{\alpha} \cos(\beta \log x) + C_2 x^{\alpha} \sin(\beta \log x)$.

Note: although the text describes mainly second order differential equation, all the ideas can be easily extended to analogous differential equation of any order.

Solve differential equation $x^2 y'' - 3xy' + 4y = 0$ with initial values $y(1) = 1$ and $y'(1) = 2$.

We assume that the solution is in the form $y(x) = x^m$. Then

$$ \begin{eqnarray} y(x) &&=&& x^m, \nonumber \\ y'(x) &&=&& m x^{m-1}, \nonumber \\ y''(x) &&=&& m (m-1) x^{m-2}.\nonumber \end{eqnarray} $$Let us use the solution within given differential equation:

$$ \begin{eqnarray} \color{red}{x^2} \cdot m (m-1) \color{red}{x^{m-2}} - 3\color{red}{x} \cdot m\color{red}{x^{m-1}} + 4 \cdot \color{red}{x^m} &&=&& 0 \nonumber \\ \color{red}{x^m} ((m^2-m) - 3m + 4) &&=&& 0 \nonumber\\ x^m (m^2- 4m + 4) &&=&& 0\nonumber\\ m &&=&& \{2, 2\}\nonumber\\ \end{eqnarray} $$The solution of differential equation is

$$ \begin{equation} \underline{y_c(x) = c_1 x^2 + c_2 x^2 \log x} \label{co:ref1}. \end{equation} $$In order to find values of $c_1$ and $c_2$ according to initial values, we need also $y_c'$:

$$ \begin{eqnarray} y_c'(x) &&=&& 2 c_1 x + 2 c_2 x \log x + c_2 x^2 \frac{1}{x} = \nonumber \\ &&=&& 2c_1 x + 2 c_2 x \log x + c_2 x. \label{co:ref2} \end{eqnarray} $$For $y(1) = 1$ and $y'(1) = 2$ we get from $(\ref{co:ref1})$ and $(\ref{co:ref2})$ $c_1 = 1$, $c_2 = 0$. The solution is then

$$ \underline{\underline{y(x) = x^2}}. $$Solve differential equation $x^2y'' - 2xy' + 2y = x^3, \hskip1em x>0$

Because the differential equation is in the form of Cauchy-Euler DE we assume that the solution is in the form of $y(x)=x^m$. Then

$$ \begin{eqnarray} y(x) &&=&& x^m, \nonumber \\ y'(x) &&=&& m x^{m-1}, \nonumber \\ y''(x) &&=&& m (m-1) x^{m-2}.\nonumber \end{eqnarray} $$Let us use the solution within associated homogeneous differential equation to obtain complementary function $y_c(x)$:

$$ \begin{eqnarray} \color{red}{x^2} \cdot m (m-1) \color{red}{x^{m-2}} - 2\color{red}{x} \cdot m\color{red}{x^{m-1}} + 2 \cdot \color{red}{x^m} &&=&& 0 \nonumber \\ \color{red}{x^m} (m(m-1) - 2m + 2) &&=&& 0 \nonumber\\ x^m (m^2- 3m + 2) &&=&& 0\nonumber\\ x^m (m- 2)(m - 1) &&=&& 0\nonumber\\ m &&=&& \{1, 2\}\nonumber\\ \end{eqnarray} $$The complementary function $y_c(x)$ (i.e. the solution of associated homogeneous differential equation) is

$$ \begin{equation} \underline{y_c(x) = c_1 x + c_2 x^2} \label{co:ref3}. \end{equation} $$We have $y_c$ but we have no $y_p$, so the solution is still incomplete. The differential
equation has no constant coefficients, so we can not use *method of
undeterminated coefficients*. We have to employ *method of variation of parameters*
to determine particular solution $y_p$.

Let us take $y_1(x)$ and $y_2(x)$ from $y_c(x)$ and form particular solution $y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)$:

$$ y_1(x) = x, \hskip1em y_2(x) = x^2 \nonumber \\ y_1'(x) = 1, \hskip1em y_2'(x) = 2x \nonumber \\ $$We have to find $u_1'(x)$ and $u_2'(x)$ as method of variation of parameters requires ($f(x) = x^3 / x^2 = x$ for differential equation in standard form):

$$ \begin{eqnarray} u_1'(x) &&=&& \frac{W_1}{W} = -\frac{x^2 x}{2x^2 - x^2} = -\frac{x^3}{x^2} = -x, \nonumber\\ u_2'(x) &&=&& \frac{W_2}{W} = \frac{x\cdot x}{x^2} = 1 \nonumber \end{eqnarray} $$After integrating $u_1'(x)$ and $u_2'(x)$ we may form particular and general solution:

$$ \begin{array}{l} \underline{y_p(x) = -\frac{x^2}{2}x + x\cdot x^2 = \frac{1}{2}x^3}\\ \underline{\underline{y(x) = c_1 x + c_2 x^2 + \frac{1}{2}x^3}}. \end{array} $$Solve differential equation $x^2y''+4xy'-4y=3$

First, let us solve associated homogeneous DE $x^2y''+4xy'-4y=0$ for $y_c$, then particular solution $y_p$.

We expect the solution in a form of $y=x^m$, then $y'=mx^{m-1},\ y''=m(m-1)x^{m-2}$.

$$ \begin{align} x^2\ m(m-1)x^{m-2} + 4x\ mx^{m-1} - 4\ x^m &= 0 \nonumber \\ x^m[m(m-1) +4m -4] &= 0 \nonumber \\ x^m(m^2+3m-4) = 0\implies m &= \{-4, 1\} \nonumber \\ \end{align} $$The complementary function is $y_c = c_1x+c_2x^{-4}$ and particular solution $y_p = A$ has to be determined and it is pretty obvious that $A=-3/4$ (test yourself that $y_p = -3/4$ satisfies given DE).

$$ \underline{\underline{y = c_1x + c_2x^{-4} - \frac 3 4}} $$