Euler-Cauchy Equation
Examine how the general solution of the Euler-Cauchy equation takes different forms depending on the sign of the discriminant of the auxiliary 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 takes different forms 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 General solution 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 with constants $a$ and $b$ and unknown function $y(x)$ in the form:
\[x^2y^{\prime\prime} + axy^{\prime} + by = 0. \label{eqn:euler_cauchy_eqn}\tag{1}\]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
\[[m(m-1) + am + b]x^m = 0\]This gives us the auxiliary equation
\[m^2 + (a-1)m + b = 0 \label{eqn:auxiliary_eqn}\tag{2}\]For $y=x^m$ to be a solution of the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$), $m$ must be a root of the auxiliary equation ($\ref{eqn:auxiliary_eqn}$).
The roots of this quadratic equation are
\[\begin{align*} m_1 &= \frac{1}{2}\left[(1-a) + \sqrt{(1-a)^2 - 4b} \right], \\ m_2 &= \frac{1}{2}\left[(1-a) - \sqrt{(1-a)^2 - 4b} \right] \end{align*}\label{eqn:m1_and_m2}\tag{3}\]Thus, the functions
\[y_1 = x^{m_1}, \quad y_2 = x^{m_2}\]are solutions to equation ($\ref{eqn:euler_cauchy_eqn}$).
As with homogeneous linear ODEs with constant coefficients, we can classify the solutions into three cases based on the discriminant $(1-a)^2 - 4b$ of the auxiliary equation ($\ref{eqn:auxiliary_eqn}$):
- $(1-a)^2 - 4b > 0$: Two distinct real roots
- $(1-a)^2 - 4b = 0$: One real double root
- $(1-a)^2 - 4b < 0$: Complex conjugate roots
General Solutions Based on the Discriminant
I. Two Distinct Real Roots $m_1$ and $m_2$
In this case, the basis for equation ($\ref{eqn:euler_cauchy_eqn}$) 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 auxiliary equation ($\ref{eqn:auxiliary_eqn}$) has a single root $m = m_1 = m_2 = \cfrac{1-a}{2}$. This gives us one solution
\[y_1 = x^{(1-a)/2}\]and the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$) becomes
\[y^{\prime\prime} + \frac{a}{x}y^{\prime} + \frac{(1-a)^2}{4x^2}y = 0 \label{eqn:standard_form}\tag{5}\]To find another linearly independent solution, we use the reduction of order method.
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 since the ratio $\frac{y_2}{y_1} = \ln x$ is not constant, $y_1$ and $y_2$ are linearly independent. The general solution 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 roots 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 to equation ($\ref{eqn:euler_cauchy_eqn}$) can be written as follows, using $x=e^{\ln x}$:
\[\begin{align*} x^{m_1} &= x^{(1-a)/2 + i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}(e^{\ln x})^{i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}e^{i(\sqrt{b - \frac{1}{4}(1-a)^2}\ln x)}, \\ x^{m_2} &= x^{(1-a)/2 - i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}(e^{\ln x})^{-i\sqrt{b - \frac{1}{4}(1-a)^2}} \\ &= x^{(1-a)/2}e^{i(-\sqrt{b - \frac{1}{4}(1-a)^2}\ln x)}. \end{align*} \tag{7}\]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
\[\begin{align*} x^{m_1} &= x^{(1-a)/2}\left[\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) + i\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right], \\ x^{m_2} &= x^{(1-a)/2}\left[\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) - i\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right] \end{align*} \tag{8}\]From these, we can derive 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 solutions are linearly independent and form a basis. By the superposition principle, the general solution is:
\[y = x^{(1-a)/2} \left[ A\cos\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) + B\sin\left(\sqrt{b - \tfrac{1}{4}(1-a)^2}\ln x \right) \right]. \label{eqn:general_sol_3}\tag{10}\]Note that the case of complex conjugate roots in Euler-Cauchy equations is of less practical importance.
Transformation to Homogeneous Linear ODEs with Constant Coefficients
The Euler-Cauchy equation can be transformed into a homogeneous linear ODE with constant coefficients through a change of variables.
With the substitution $x = e^t$, we have:
\[\frac{d}{dx} = \frac{1}{x}\frac{d}{dt}, \quad \frac{d^2}{dx^2} = \frac{1}{x^2}\left(\frac{d^2}{dt^2} - \frac{d}{dt} \right)\]This transforms the Euler-Cauchy equation ($\ref{eqn:euler_cauchy_eqn}$) into:
\[y^{\prime\prime}(t) + (a-1)y^{\prime}(t) + by(t) = 0. \label{eqn:substituted}\tag{11}\]We can solve equation ($\ref{eqn:substituted}$) using methods for homogeneous linear ODEs with constant coefficients, and then substitute $t = \ln{x}$ to obtain the solution in terms of $x$. This approach yields the same results as discussed above.