Euler-Cauchy Equation

TL;DR

• Euler-Cauchy equation: $x^2y^{\prime\prime} + axy^{\prime} + by = 0$
• Auxiliary equation: $m^2 + (a-1)m + b = 0$
• The general solution can be categorized into three cases depending on the sign of the discriminant $(1-a)^2 - 4b$ of the auxiliary equation, as shown in the table

Case	Roots of auxiliary equation	Basis for Euler-Cauchy equation solutions	General solution of Euler-Cauchy equation
I	Distinct real roots $m_1$, $m_2$	$x^{m_1}$, $x^{m_2}$	$y = c_1 x^{m_1} + c_2 x^{m_2}$
II	Real double root $m = \cfrac{1-a}{2}$	$x^{(1-a)/2}$, $x^{(1-a)/2}\ln{x}$	$y = (c_1 + c_2 \ln x)x^m$
III	Complex conjugate roots $m_1 = \cfrac{1}{2}(1-a) + i\omega$, $m_2 = \cfrac{1}{2}(1-a) - i\omega$	$x^{(1-a)/2}\cos{(\omega \ln{x})}$, $x^{(1-a)/2}\sin{(\omega \ln{x})}$	$y = x^{(1-a)/2}[A\cos{(\omega \ln{x})} + B\sin{(\omega \ln{x})}]$

Prerequisites

• Homogeneous Linear ODEs of Second Order
• Homogeneous Linear ODEs with Constant Coefficients
• Euler’s formula

Auxiliary Equation

The Euler-Cauchy equation is a differential equation of the form

\[x^2y^{\prime\prime} + axy^{\prime} + by = 0 \label{eqn:euler_cauchy_eqn}\tag{1}\]

where $a$ and $b$ are constants and $y(x)$ is the unknown function. Substituting

\[y=x^m, \qquad y^{\prime}=mx^{m-1}, \qquad y^{\prime\prime}=m(m-1)x^{m-2}\]

into equation ($\ref{eqn:euler_cauchy_eqn}$), we get

\[x^2m(m-1)x^{m-2} + axmx^{m-1} + bx^m = 0,\]

which simplifies to

\[[m(m-1) + am + b]x^m = 0\]

From this, we obtain the auxiliary equation

\[m^2 + (a-1)m + b = 0 \label{eqn:auxiliary_eqn}\tag{2}\]

The necessary and sufficient condition for $y=x^m$ to be a solution of the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$) is that $m$ is a solution of the auxiliary equation ($\ref{eqn:auxiliary_eqn}$).

Solving the quadratic equation ($\ref{eqn:auxiliary_eqn}$), we get

\[\begin{align*} m_1 &= \frac{1}{2}\left[(1-a) + \sqrt{(1-a)^2 - 4b} \right], \\ m_2 &= \frac{1}{2}\left[(1-a) - \sqrt{(1-a)^2 - 4b} \right] \end{align*}\label{eqn:m1_and_m2}\tag{3}\]

Thus, the two functions

\[y_1 = x^{m_1}, \quad y_2 = x^{m_2}\]

are solutions of equation ($\ref{eqn:euler_cauchy_eqn}$).

As with homogeneous linear ODEs with constant coefficients, we can divide the cases into three categories based on the sign of the discriminant $(1-a)^2 - 4b$ of the auxiliary equation ($\ref{eqn:auxiliary_eqn}$):

• $(1-a)^2 - 4b > 0$: Two distinct real roots
• $(1-a)^2 - 4b = 0$: Real double root
• $(1-a)^2 - 4b < 0$: Complex conjugate roots

General Solution Forms Based on the Discriminant Sign

I. Two Distinct Real Roots $m_1$ and $m_2$

In this case, the basis of solutions for equation ($\ref{eqn:euler_cauchy_eqn}$) on any interval is

\[y_1 = x^{m_1}, \quad y_2 = x^{m_2}\]

and the general solution is

\[y = c_1 x^{m_1} + c_2 x^{m_2} \label{eqn:general_sol_1}\tag{4}\]

II. Real Double Root $m = \cfrac{1-a}{2}$

When $(1-a)^2 - 4b = 0$, i.e., $b=\cfrac{(1-a)^2}{4}$, the quadratic equation ($\ref{eqn:auxiliary_eqn}$) has only one solution $m = m_1 = m_2 = \cfrac{1-a}{2}$. Thus, one solution of the form $y = x^m$ is

\[y_1 = x^{(1-a)/2}\]

and the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$) becomes

\[y^{\prime\prime} + \frac{a}{x}y^{\prime} + \frac{(1-a)^2}{4x^2}y = 0 \label{eqn:standard_form}\tag{5}\]

Let’s find another linearly independent solution $y_2$ using reduction of order.

Setting $y_2=uy_1$, we get

\[u = \int U, \qquad U = \frac{1}{y_1^2}\exp\left(-\int \frac{a}{x}\ dx \right)\]

Since $\exp \left(-\int \cfrac{a}{x}\ dx \right) = \exp (-a\ln x) = \exp(\ln{x^{-a}}) = x^{-a}$,

\[U = \frac{x^{-a}}{y_1^2} = \frac{x^{-a}}{x^{(1-a)}} = \frac{1}{x}\]

Integrating, we get $u = \ln x$.

Therefore, $y_2 = uy_1 = y_1 \ln x$, and $y_1$ and $y_2$ are linearly independent since their ratio is not constant. The general solution corresponding to the basis $y_1$ and $y_2$ is

\[y = (c_1 + c_2 \ln x)x^m \label{eqn:general_sol_2}\tag{6}\]

III. Complex Conjugate Roots

In this case, the solutions of the auxiliary equation ($\ref{eqn:auxiliary_eqn}$) are $m = \cfrac{1}{2}(1-a) \pm i\sqrt{b - \frac{1}{4}(1-a)^2}$, and the corresponding complex solutions of equation ($\ref{eqn:euler_cauchy_eqn}$) can be written using $x=e^{\ln x}$ as follows:

\[\begin{align*} x^{m_1} &= x^{(1-a)/2 + i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}(e^{\ln x})^{i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}e^{i(\sqrt{b - \frac{1}{4}(1-a)^2}\ln x)}, \\ x^{m_2} &= x^{(1-a)/2 - i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}(e^{\ln x})^{-i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}e^{i(-\sqrt{b - \frac{1}{4}(1-a)^2}\ln x)}. \end{align*} \tag{7}\]

Setting $t=\sqrt{b - \frac{1}{4}(1-a)^2}\ln x$ and using Euler’s formula $e^{it} = \cos{t} + i\sin{t}$, we get

\[\begin{align*} x^{m_1} &= x^{(1-a)/2}\left[\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) + i\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right], \\ x^{m_1} &= x^{(1-a)/2}\left[\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) - i\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right] \end{align*} \tag{8}\]

From these, we obtain the following two real solutions:

\[\begin{align*} \frac{x^{m_1} + x^{m_2}}{2} &= x^{(1-a)/2}\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right), \\ \frac{x^{m_1} - x^{m_2}}{2i} &= x^{(1-a)/2}\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \end{align*} \tag{9}\]

Since their ratio $\cos\left(\sqrt{b - \frac{1}{4}(1-a)^2}\ln x \right)$ is not constant, these two solutions are linearly independent and thus form a basis for the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$) by the principle of superposition. This gives us the following real general solution:

\[y = x^{(1-a)/2} \left[ A\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) + B\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right]. \label{eqn:general_sol_3}\tag{10}\]

However, the case where the auxiliary equation has complex conjugate roots is not of great practical importance in Euler-Cauchy equations.