Preface
This section presents a class of variable coefficient equations that admit closed form solutions---the Euler equations (also known as equidimensional equations).
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Glossary
Euler equations
Linear differential equations with variable coefficients rarely have explicit solutions expressed via familiar functions. However, there is a special class of variable coefficient linear differential equations where explicit solutions are not difficult to obtain. They were discovered by Leonhard Euler (1707--1783).
The n-th order homogeneous Euler differential equation is an equation of the form
This equations are also known Cauchy--Euler equation or Euler--Cauchy equations. The most common term for these equations is the equidimensional equation because products of powers of x and corresponding derivatives, \( x^k y^{(k)} \ k = 0,1,2,\ldots , \) have the same units as the function y(x). This keeps the coefficients \( a_0 , a_1 , \ldots , a_n \) dimensionless. Since the leading coefficient in equation \eqref{EqEuler.2} is 0 when x = 0, this point is a singular point for Euler's equation; this forces us to consider Euler's differential equation in domains not containing the origin: either for x > 0 or x < 0. As we will see in the following examples, initial value problems with initial conditions specified at the origin are ill-posed: their solutions either do not exist or have infinite many solutions.
\begin{equation} \label{EqEuler.1}
a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \cdots + a_1 x\,y' + a_0 y =0 \qquad (a_n \ne 0) ,
\end{equation}
where \( a_0 , a_1 , \ldots , a_n \) are some real constants and \( y^{(k)} (x) = {\text d}^k y/{\text d}x^k \) is the kth derivative. The general inhomogeneous Euler equation is
\begin{equation} \label{EqEuler.2}
a_n x^n y^{(n)} + a_{n-1} x^{n-1} y^{(n-1)} + \cdots + a_1 x\,y' + a_0 y = f(x) \qquad (a_n \ne 0) ,
\end{equation}
where f(x) is a given function.
Euler's equations can be generalized as
\begin{equation} \label{EqEuler.3}
a_n \left( \alpha x + \beta \right)^n y^{(n)} + a_{n-1} \left( \alpha x + \beta \right)^{n-1} y^{(n-1)} + \cdots + a_1 \left( \alpha x + \beta \right) y' + a_0 y =0 \qquad (a_n \ne 0) ,
\end{equation}
with some real numbers α and β. Upon changing independent variable t = αx + β equidimensional equation \eqref{EqEuler.3} can be transferred into standard Euler's equation \eqref{EqEuler.1} having singular point at the origin instead of −β/α.
When n = 2, Euler's homogeneous equation can be written as
\begin{equation} \label{EqEuler.4}
a\,x^2 y'' + b\,x\, y' + c\,y =0 \qquad (x > 0),
\end{equation}
for some real constants \( a, b, c . \) Euler's equation can be solved explicitly by substitution t = ln|x| > 0, or
\[
x = \begin{cases}
e^t , & \ \mbox{ if } \ x > 0, \\
e^{-t} , & \ \mbox{ if } \ x < 0.
\end{cases}
\]
Then
\begin{align*}
\frac{{\text d}y}{{\text d} x} &= \frac{{\text d}y}{{\text d} t} \, \frac{{\text d}t}{{\text d} x} = \frac{{\text d}y}{{\text d} t} \,\frac{1}{x} ,
\\
\frac{{\text d}^2 y}{{\text d} x^2} &= \frac{{\text d}}{{\text d} x} \left( \frac{{\text d}y}{{\text d} t} \,\frac{1}{x} \right) = \frac{{\text d}^2 y}{{\text d} t^2} \,\frac{1}{x^2} - \frac{{\text d}y}{{\text d} t} \,\frac{1}{x^2} .
\end{align*}
Substituting these expressions into Euler's homogeneous equation \eqref{EqEuler.3}, we get
\[
a\,x^2 \left[ \frac{{\text d}^2 y}{{\text d} t^2} \,\frac{1}{x^2} - \frac{{\text d}y}{{\text d} t} \,\frac{1}{x^2} \right] + b\,x \left[ \frac{{\text d}y}{{\text d} t} \,\frac{1}{x} \right] + c\, y = 0.
\]
Upon canceling the common multiple x², we have constant coefficient differential equation
\[
a\,\frac{{\text d}^2 y}{{\text d} t^2} + \left( b-a \right) \frac{{\text d}y}{{\text d} t} + c\, y = 0.
\]
Thus, we get a constant coefficient differential equation for which we know everything from the previous discussion in Part IV: of this tutorial. The characteristic equation for the above constant coefficient differential equation
\begin{equation} \label{EqEuler.5}
a\,\nu^2 + \left( b - a \right) \nu + c = 0
\end{equation}
is called the indicial equation for the Euler homogeneous equation \eqref{EqEuler.3}.
Euler's equation \eqrefl{EqEuler.1} can also be solved by trial solution y = xν. Upon substituting it into the differential equation \eqrefl{EqEuler.3}, we get
\[
a\, \nu \left( \nu-1 \right) x^2 x^{\nu -2} + b\,\nu\,x^{\nu} + c\,x^{\nu} =0 .
\]
Canceling the common multiple xν, we get
the indicial equation
\[
P(\nu ) = a\, \nu\left( \nu -1 \right) + b\, \nu + c = 0.
\]
If roots ν1 and ν2 of the indicial equation are distinct real numbers, then the general solution of Euler's equation is given by
\[
y(x) = C_1 |x|^{\nu_1} + C_2 |x|^{\nu_2}
\]
for some arbitrary real constants C1 and C2. If the roots ν1 and ν2 are equal (ν1 = ν2), then the general solution of Euler's equation is given by
\[
y(x) = C_1 |x|^{\nu_1} + C_2 |x|^{\nu_1} \,\ln |x| .
\]
If roots ν1 and ν2 of the indicial equation are complex numbers, then the general solution of Euler's equation is
\[
y(x) = \left[ C_1 \cos \left( \beta\,\ln |x| \right) + C_2 \sin \left( \beta\,\ln |x| \right) \right] |x|^{\alpha} ,
\]
where
\[
\alpha = \frac{a-b}{2a} \qquad\mbox{and} \qquad \beta = \frac{\sqrt{4ac - (b-a)^2}}{2a} .
\]
We summarize properties of this equation in the following table.
| Solutions of the indicial equation | Fundamental set of solutions |
|---|---|
| Simple roots m1, m2 | \( y_1 = x^{m_1} , \ y_2 = x^{m_2} , \ldots \) |
| Double roots \( m_1 = m_2 \) | \( y_1 = x^{m_1} , \ y_2 = x^{m_1} \ln x \) |
| Triple roots \( m_1 = m_2 = m_3 \) | \( y_1 = x^{m_1} , \ y_2 = x^{m_1} \ln x , \ y_3 = x^{m_1} \left( \ln x \right)^2 \) |
| Complex roots \( m_1 = \overline{m_2} = \alpha + \beta {\bf j} \) | \( y_1 = x^{m_1} \cos \left( \beta\,\ln x \right) , \ y_2 = x^{m_1} \sin \left( \beta\,\ln x \right) \) |
Example 1: Consider the differential equation
\[
\frac{{\text d}^2 y}{{\text d} x^2} + \frac{2}{x}\,\frac{{\text d}y}{{\text d}x} - \frac{6}{x^2} \, y = 0 .
\]
The corresponding indicial equation
\[
\nu^2 + \nu -6 =0
\]
has two distinct real roots ν1 = 3 and ν = −2. Therefore, the given Euler equation has the general solution:
\[
y(x) = C_1 x^3 + C_2 \frac{1}{x^2} ,
\]
where C1 and C2 are arbitrary constants.
■ Example 2: Consider the initial value problem
\[
3\,x^2 y'' -7\,x\,y' + 7\,y = 0, \qquad y(0) = a, \quad y' (0) = b,
\]
where 𝑎 and b are some real numbers. Note that both initial conditions are specified at the singular point x = 0. Substituting for y(x) = xm, we get an indicial equation
\[
3 \left( m^2 - m \right) -7\,m + 7 = 0 .
\]
It has two distinct real roots m = 1 and m = 7/3. Therefore, the given Euler equation has the general solution:
\[
y(x) = C_1 x^{7/3} + C_2 \, x ,
\]
where C1 and C2 are arbitrary constants.
To satisfy the initial conditions, we have to solve the system of equations
\[
C_1 0 + c_2 0 = a, \qquad C_2 = b .
\]
So 𝑎 must be zero and the given initial value problem has one-parameter set of solutions
\[
y (x) = C_1 x^{7/3} + b\, x .
\]
Note that if the arbitrary real constant C1 ≠ 0, its solution is not analytic.
■ Example 3: Consider the differential equation
\[
\frac{{\text d}^2 y}{{\text d} x^2} - \frac{3}{x}\,\frac{{\text d}y}{{\text d}x} + \frac{4}{x^2} \, y = 0 .
\]
The corresponding indicial equation
\[
\nu^2 -4\, \nu +4 =0
\]
has one double root ν = 2. Therefore, the given Euler equation has the general solution:
\[
y(x) = C_1 |x|^2 + C_2 |x|^2 \,\ln |x| ,
\]
where C1 and C2 are arbitrary constants.
■ Example 4: Consider the differential equation
\[
\frac{{\text d}^2 y}{{\text d} x^2} - \frac{1}{x}\,\frac{{\text d}y}{{\text d}x} + \frac{5}{x^2} \, y = 0 .
\]
The corresponding indicial equation
\[
\nu^2 -2\, \nu +5 =0
\]
has two complex conjugate roots ν± = -1 ± 2. The the general solution of the given Euler's equation is
\[
y(x) = \left[ C_1 \cos \left( 2\ln |x| \right) + C_2 \sin \left( 2\ln |x| \right) \right] |x|^{-1} = \left[ C_1 \cos \left( \ln x^2 \right) + C_2 \sin \left( \ln x^2 \right) \right] |x|^{-1} ,
\]
where C1 and C2 are arbitrary constants.
■ Example 5: ■
Singularities of Euler's Equation
There is generally another singularity at the point at infinity in the complex plane ℂ, whose nature is found by mapping the point at infinity to the origin via substitution x = 1/t. The change of variable of differentiation has the effect
\begin{align*}
\frac{{\text d}y}{{\text d}x} &= -t^2 \frac{{\text d}u(t)}{{\text d}t} = -t^2 \dot{u}
\\
\frac{{\text d}^2 y}{{\text d}x^2} &= t^4 \frac{{\text d}^2 u(t)}{{\text d}t^2} + 2 t^3 \frac{{\text d}u(t)}{{\text d}t} = t^4 \ddot{u} + 2 t^3 \dot{u} ,
\end{align*}
where u(t) = y(t-1). Then the second order Euler's equation \eqref{EqEuler.4} becomes
\begin{equation} \label{EqInfty.1}
a t^2 \frac{{\text d}^2 u(t)}{{\text d}t^2} + \left( 2a -b \right) t\, \frac{{\text d}u(t)}{{\text d}t} + c\, u = 0,
\end{equation}
and the singularity at infinity is seen to be regular. For a regular singular point at infinity, the approximation is written in terms of an inverse power x−1 with the indicial equation
\begin{equation} \label{EqInfty.2}
a\lambda \left( \lambda +1 \right) - b\,\lambda + c = 0 .
\end{equation}
Example 6: We consider several typical cases and analyze the behavior of solutions to Euler's equations at infinity.
Example 6A: Euler's equation
\[
x^2 y'' + 5x\,y' + 3\,y = 0
\]
has the general solution
\[
y = C_1 x^{-1} + C_2 x^{-3} ,
\]
with some arbitrary constants C1 and C2 because its indicial equation ν² + 4ν + 3 = 0 has two real roots ν = −1 and ν = −3.
For regular singular point at infinity, we make substitution x = t−1, which leads to
\[
t^2 \ddot{u} -4\,\dot{u} + 3\, u = 0.
\]
Its indicial equation λ² &mimus;4λ + 3 = 0 has two real roots λ = 1 and λ = 3. Therefore, at infinity the solution behaves as
\[
u(t) = C_1 t + C_2 t^3 \qquad \Longrightarrow \qquad y = C_1 x^{-1} + C_2 x^{-3} .
\]
Example 6B: Euler's equation
\[
x^2 y'' + 7x\,y' + 9\,y = 0
\]
has the general solution
\[
y (x) = C_1 x^{-3} + C_2 x^{-3} \ln x .
\]
The corresponding indicial equation ν² + 6ν + 9 = (ν +3)² = 0 has one double root ν = −3.
Upon substitution x = t−1, we obtain
\[
t^2 \ddot{u} -5\,\dot{u} + 9\,u = 0 , \qquad u(t) = y \left( x^{-1} \right) .
\]
The corresponding indicial equation
\[
\lambda \left( \lambda + 1 \right) - 7\,\lambda + 9 = 0
\]
has one double root λ = 3, and the general solution around infinity has the form
\[
u(t) = C_1 t^3 + C_2 t^3 \ln t \qquad Longrightarrow \qquad y(x) = u\left( x^{-1}\right) = C_1 x^{-3} + C_2 x^{-3} \ln x .
\]
Example 6C: The inicial equation ν² + 4ν + 13 = 0 for Euler's equation
\[
x^2 y'' +5x\,y' + 13\,y = 0
\]
has two complex conjugate roots ν = −2 ±3j, where j is the unit vector in the positive vertical direction on the complex plane ℂ, with j² = −1. Hence its general solution becomes
\[
y (x) = C_1 x^{-2} \cos \left( 3\ln x \right) + C_2 x^{-2} \sin \left( 3\ln x \right) .
\]
Upon transformation x = t−1, we obtain
\[
t^2 \ddot{u} + t \left( 2 - 5 \right) \dot{u} + 13\, u = 0 .
\]
The corresponding indicial equation
\[
\lambda \left( \lambda + 1 \right) - 5 \lambda + 13 =0
\]
has two complex conjugate roots λ = −2 ±3j. So its general solution around infinity is
\[
y(x) = u\left( x^{-1} \right) = C_1 t^2 \cos \left( 3 \ln t \right) + C_2 t^2 \sin \left( 3 \ln t \right) = C_1 x^{-2} \cos \left( 3\ln x \right) + C_2 x^{-2} \sin \left( 3\ln x \right) .
\]
■
Factorization of Euler's Equation
\begin{equation} \label{EqEuler.6}
L \left[ x,\texttt{D} \right] = x^{-m-n} \texttt{D} x^m \,\texttt{D} \left[
\left( \cdot \right) \right] = \texttt{D}^2 + \frac{2n+m}{x}\,\texttt{D}
+ \frac{n(m+n-1)}{x^2}\,\texttt{I} ,
\end{equation}
where m, n are some real numbers, and
\[
\texttt{D} = \frac{\text d}{{\text d} x} \qquad \mbox{and} \qquad \texttt{I} = \texttt{D}^0
\]
are the derivative operator and the identical operator, respectively. It is also sometimes convenient to eliminate fractions and consider a similar operator:
\begin{equation} \label{EqEuler.7}
L \left[ x,\texttt{D} \right] = x^{-m-n} \texttt{D} x^m \,\texttt{D} \left[
\left( \cdot \right) \right] = \texttt{D}^2 + \frac{2n+m}{x}\,\texttt{D}
+ \frac{n(m+n-1)}{x^2}\,\texttt{I} ,
\end{equation}
where m, n are some real numbers, and
\[
\texttt{D} = \frac{\text d}{{\text d} x} \qquad \mbox{and} \qquad \texttt{I} = \texttt{D}^0
\]
are the derivative operator and the identical operator, respectively. It is also sometimes convenient to eliminate fractions and consider a similar operator:
\begin{equation} \label{EqEuler.8}
L \left[ x,\texttt{D} \right] = x^{-m-n} \texttt{D} x^m \,\texttt{D} \left[
\left( \cdot \right) \right] = \texttt{D}^2 + \frac{2n+m}{x}\,\texttt{D}
+ \frac{n(m+n-1)}{x^2}\,\texttt{I} ,
\end{equation}
where m, n are some real numbers, and
\[
\texttt{D} = \frac{\text d}{{\text d} x} \qquad \mbox{and} \qquad \texttt{I} = \texttt{D}^0
\]
are the derivative operator and the identical operator, respectively. It is also sometimes convenient to eliminate fractions and consider a similar operator:
\begin{equation} \label{EqEuler.9}
x^2 L \left[ x, \texttt{D} \right] = x^{2-m-n} \texttt{D} x^m \,\texttt{D} \left[
x^n \left( \cdot \right) \right] = x^2 \texttt{D}^2 + \left( 2n+m \right) \texttt{D} + n(m+n-1)\,\texttt{I} .
\end{equation}
Simplify[D[(x^m)*D[f[x]*x^n, x], x]]
x^(-2 + m +
n) (n (-1 + m + n) f[x] +
x ((m + 2 n) Derivative[1][f][x] + x (f^\[Prime]\[Prime])[x]))
General equidimensional operator of the second order also can be factorized:
\begin{equation} \label{EqEuler.10}
\left( \alpha x + \beta \right)^2 L \left[ x, \texttt{D} \right] = \left( \alpha x + \beta \right)^{2-m-n} \texttt{D} \left( \alpha x + \beta \right)^m \,\texttt{D} \left( \alpha x + \beta \right)^n \left[
\left( \cdot \right) \right] = \left( \alpha x + \beta \right)^2 \texttt{D}^2 + \alpha \left( \alpha x + \beta \right) \left( 2n+m \right) \texttt{D} + \alpha^2 \left( \alpha x + \beta \right)^2 n(m+n-1)\,\texttt{I} .
\end{equation}
We want to express an arbitrary Euler's operator
\[
L \left[ x, \texttt{D} \right] = \frac{{\text d}^2}{{\text d}x^2} + \frac{p}{x}\,\frac{{\text d}}{{\text d}x} + \frac{q}{x^2} = \texttt{D}^2 + \frac{p}{x}\, \texttt{D} + \frac{q}{x^2}\, \texttt{I} ,
\]
where p and q are some real numbers, in the
factorized form. Therefore, we need to solve the system of algebraic equations:
\[
\begin{cases}
p &= m+2n , \\
q &= n \left( n+m-1 \right) ,
\end{cases} \qquad \Longrightarrow \qquad
\begin{cases}
n&= \frac{1}{2} \left[ p-1 \pm \sqrt{(1-p)^2 - 4\,q} \right] , \\
m &= 1 \pm \sqrt{(p-1)^2 - 4\,q} .
\end{cases}
\]
Solve[{p == 2*n + m, q == n*(n + m - 1)}, {n, m}]
{{n -> 1/2 (-1 + p - Sqrt[1 - 2 p + p^2 - 4 q]),
m -> 1 + Sqrt[1 - 2 p + p^2 - 4 q]}, {n ->
1/2 (-1 + p + Sqrt[1 - 2 p + p^2 - 4 q]),
m -> 1 - Sqrt[1 - 2 p + p^2 - 4 q]}}
Therefore, the Euler operator admits factorization only when
\begin{equation} \label{EqEuler.11}
(p-1)^2 \ge 4\,q .
\end{equation}
Otherwise, the factorization system of algebraic equations leads to complex roots that we do not consider. Note that factorization \eqref{EqEuler.8} or \eqref{EqEuler.10} is not unique. The value n in these formulas is one of the roots of the indicial equation.
Example 7: Consider the singular differential operator
\[
L \left[ x,\texttt{D} \right] =
\frac{{\text d}^2}{{\text d} x^2} + \frac{5}{x}\,\frac{{\text d}}{{\text d}x} = \texttt{D}^2 + \frac{5}{x}\,\texttt{D} .
\]
This operator admits factorization
\[
L \left[ x,\texttt{D} \right] = x^{-5} \texttt{D}\, x^5 \,\texttt{D} .
\]
To find its inverse, we need to solve the differential equation
\[
L \left[ x,\texttt{D} \right] y = w \qquad \Longrightarrow \qquad
x^5 \texttt{D}\, y = \left( x^5 \texttt{D}\, y \right)_{x= x_0} + \int_{x_0}^x t^5 w(t)\,{\text d} t .
\]
Next integration yields
\[
L^{-1} \left[ x,\texttt{D} \right] w = y\left( x_0 \right) + y' \left( x_0 \right) \frac{1}{4} \left( x_0 - \frac{x_0^5}{x^4} \right) + \int_{x_0}^x s^{-5} {\text d} s \int_{x_0}^s {\text d} t \, t^5 w(t) .
\]
The double integral can be reduced to one integral by interchanging the order of integration
\[
L^{-1} \left[ x,\texttt{D} \right] w = y\left( x_0 \right) + y' \left( x_0 \right) \frac{1}{4} \left( x_0 - \frac{x_0^5}{x^4} \right) - \frac{1}{4} \int_{x_0}^x {\text d} t \left( t^5 x^{-4} - t \right) w(t) .
\]
■ Example 8: Let us consider Euler's equation
\[
x^2 y'' -x\,y' + y=f(x) ,
\]
where primes correspond to derivatives: y' = dy/dx. Its indicial equation r² −2r + 1 = 0 has one double root r = 1. Solving the factorization system of equations
\[
m + 2n = -1, \qquad n \left( m+n-1 \right) = 1 ,
\]
Solve[{p + 2*q == -1, q*(p + q - 1) == 1}, {p, q}]
{{p -> 1, q -> -1}}
we see that we have the only one factorization for the given Euler's equation:
\[
x^2 \texttt{D}^2 - x\,\texttt{D} + \texttt{I} = x^2 \texttt{D}\, x\, \texttt{D}\,x \, \texttt{I} .
\]
Dividing by x², we get
\[
\texttt{D}\, x\, \texttt{D}\,x \,y = x^{-2} f(x) .
\]
Integration yields
\[
x\, \texttt{D}\,x \,y = \int x^{-2} f(x) \,{\text d}x + c_1 \qquad \Longrightarrow \qquad \texttt{D}\,x \,y = x^{-1} \int x^{-2} f(x) \,{\text d}x + c_1 x^{-1} .
\]
Next integration provides us the general solution in any domain not containing the origin.
■
Nonhomogeneous Euler Equations
\[
\frac{{\text d}^2 y}{{\text d} x^2} + \frac{p}{x}\,
\frac{{\text d} y}{{\text d}x} + \frac{q}{x^2}\,y = g(x) ,
\]
with some real constants p, q, and given function
g(x). We denote the left-hand side of the above equation by
\( L \left[ y(x) \right] = y'' (x) + p\, y' (x)/x + q\,y(x)/x^2 , \) the linear differential operator with variable coefficients.
Since it contains the reciprocal of x, the origin x = 0 is a
singular point for the Euler equation. This means that we can obtain its solutions either for negative values of the argument, x < 0, or for positive
values x > 0. The initial condition can be chosen at any point other than the origin. Therefore, we consider the Euler equations only for positive values x > 0.
- Dettman, J.W., Power Series Solutions of Ordinary Differential Equations, The American Mathematical Monthly, 1967, Vol. 74, No. 3, pp. 428--430.
- Grigorieva, E., Methods of Solving Sequence and Series Problems, Birkhäuser; 1st ed. 2016.
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