Cauchy–Euler equation

The equation

Let y⁽ⁿ⁾(x) be the nth derivative of the unknown function y(x). Then a Cauchy–Euler equation of order n has the form $${\displaystyle a_n{x}^n{y}^{(n)}(x)+a_{n-1}{x}^{n-1}{y}^{(n-1)}(x)+\cdots +a_{0}y(x)=0.}$$

The substitution ${\displaystyle x=e^u}$ (that is, ${\displaystyle u=\ln (x)}$; for ${\displaystyle x<0}$, in which one might replace all instances of ${\displaystyle x}$ by ${\displaystyle |x|}$, extending the solution's domain to ${\displaystyle \mathbb{R}\setminus \{0\}}$) can be used to reduce this equation to a linear differential equation with constant coefficients. Alternatively, the trial solution ${\displaystyle y=x^m}$ can be used to solve the equation directly, yielding the basic solutions.

Second order – solving through trial solution

The most common Cauchy–Euler equation is the second-order equation, which appears in a number of physics and engineering applications, such as when solving Laplace's equation in polar coordinates. The second order Cauchy–Euler equation is

$${\displaystyle x^2\frac{d^{2}y}{d{x}^2}+ax\frac{dy}{dx}+by=0.}$$

We assume a trial solution $${\displaystyle y=x^m.}$$

Differentiating gives $${\displaystyle \frac{dy}{dx}=m{x}^{m-1}}$$ and $${\displaystyle \frac{d^{2}y}{d{x}^2}=m\left(m-1\right)x^{m-2}.}$$

Substituting into the original equation leads to requiring that $${\displaystyle x^2\left(m\left(m-1\right)x^{m-2}\right)+ax\left(m{x}^{m-1}\right)+b\left(x^m\right)=0}$$

Rearranging and factoring gives the indicial equation $${\displaystyle m^2+\left(a-1\right)m+b=0.}$$

We then solve for m. There are three cases of interest:

• Case 1 of two distinct roots, m₁ and m₂;
• Case 2 of one real repeated root, m;
• Case 3 of complex roots, α ± βi.

In case 1, the solution is $${\displaystyle y=c_1{x}^{m_1}+c_2{x}^{m_2}}$$

In case 2, the solution is $${\displaystyle y=c_1{x}^m\ln (x)+c_2{x}^m}$$

To get to this solution, the method of reduction of order must be applied, after having found one solution y = x^m.

In case 3, the solution is $${\displaystyle y=c_1{x}^{\alpha }\cos (\beta \ln (x))+c_2{x}^{\alpha }\sin (\beta \ln (x))}$$ $${\displaystyle \alpha =\mathrm{Re}(m)}$$ $${\displaystyle \beta =\mathrm{Im}(m)}$$

For ${\displaystyle c_1,c_2\in \mathbb{R}}$.

This form of the solution is derived by setting x = e^t and using Euler's formula.

Second order – solution through change of variables

$${\displaystyle x^2\frac{d^{2}y}{d{x}^2}+ax\frac{dy}{dx}+by=0}$$

We operate the variable substitution defined by

$${\displaystyle t=\ln (x).}$$ $${\displaystyle y(x)=\phi (\ln (x))=\phi (t).}$$

Differentiating gives $${\displaystyle \frac{dy}{dx}=\frac{1}{x}\frac{d\phi }{dt}}$$ $${\displaystyle \frac{d^{2}y}{d{x}^2}=\frac{1}{x^2}\left(\frac{d^2\phi }{d{t}^2}-\frac{d\phi }{dt}\right).}$$

Substituting ${\displaystyle \phi (t)}$ the differential equation becomes $${\displaystyle \frac{d^2\phi }{d{t}^2}+(a-1)\frac{d\phi }{dt}+b\phi =0.}$$

This equation in ${\displaystyle \phi (t)}$ is solved via its characteristic polynomial $${\displaystyle {\lambda }^2+(a-1)\lambda +b=0.}$$

Now let ${\displaystyle {\lambda }_1}$ and ${\displaystyle {\lambda }_2}$ denote the two roots of this polynomial. We analyze the case in which there are distinct roots and the case in which there is a repeated root:

If the roots are distinct, the general solution is $${\displaystyle \phi (t)=c_1{e}^{{\lambda }_{1}t}+c_2{e}^{{\lambda }_{2}t},}$$ where the exponentials may be complex.

If the roots are equal, the general solution is $${\displaystyle \phi (t)=c_1{e}^{{\lambda }_{1}t}+c_{2}t{e}^{{\lambda }_{1}t}.}$$

In both cases, the solution ${\displaystyle y(x)}$ can be found by setting ${\displaystyle t=\ln (x)}$.

Hence, in the first case, $${\displaystyle y(x)=c_1{x}^{{\lambda }_1}+c_2{x}^{{\lambda }_2},}$$ and in the second case, $${\displaystyle y(x)=c_1{x}^{{\lambda }_1}+c_2\ln (x)x^{{\lambda }_1}.}$$

Second order - solution using differential operators

Observe that we can write the second-order Cauchy-Euler equation in terms of a linear differential operator ${\displaystyle L}$ as $${\displaystyle Ly=(x^2{D}^2+axD+bI)y=0,}$$ where ${\displaystyle D=\frac{d}{dx}}$ and ${\displaystyle I}$ is the identity operator.

We express the above operator as a polynomial in ${\displaystyle xD}$, rather than ${\displaystyle D}$. By the product rule, $${\displaystyle (xD{)}^2=xD(xD)=x(D+x{D}^2)=x^2{D}^2+xD.}$$ So, $${\displaystyle L=(xD{)}^2+(a-1)(xD)+bI.}$$

We can then use the quadratic formula to factor this operator into linear terms. More specifically, let ${\displaystyle {\lambda }_1,{\lambda }_2}$ denote the (possibly equal) values of $${\displaystyle -\frac{a-1}{2}\pm \frac{1}{2}\sqrt{(a-1{)}^2-4b}.}$$ Then, $${\displaystyle L=(xD-{\lambda }_{1}I)(xD-{\lambda }_{2}I).}$$

It can be seen that these factors commute, that is ${\displaystyle (xD-{\lambda }_{1}I)(xD-{\lambda }_{2}I)=(xD-{\lambda }_{2}I)(xD-{\lambda }_{1}I)}$. Hence, if ${\displaystyle {\lambda }_1\ne {\lambda }_2}$, the solution to ${\displaystyle Ly=0}$ is a linear combination of the solutions to each of ${\displaystyle (xD-{\lambda }_{1}I)y=0}$ and ${\displaystyle (xD-{\lambda }_{2}I)y=0}$, which can be solved by separation of variables.

Indeed, with ${\displaystyle i\in \{1,2\}}$, we have ${\displaystyle (xD-{\lambda }_{i}I)y=x\frac{dy}{dx}-{\lambda }_{i}y=0}$. So, $${\displaystyle \begin{array}{cc}x\frac{dy}{dx}& ={\lambda }_{i}y\\ \int \frac{1}{y}dy& ={\lambda }_i\int \frac{1}{x}dx\\ \ln y& ={\lambda }_i\ln x+C\\ y& =c_i{e}^{{\lambda }_i\ln x}=c_i{x}^{{\lambda }_i}.\end{array}}$$ Thus, the general solution is ${\displaystyle y=c_1{x}^{{\lambda }_1}+c_2{x}^{{\lambda }_2}}$.

If ${\displaystyle \lambda ={\lambda }_1={\lambda }_2}$, then we instead need to consider the solution of ${\displaystyle (xD-\lambda I{)}^{2}y=0}$. Let ${\displaystyle z=(xD-\lambda I)y}$, so that we can write $${\displaystyle (xD-\lambda I{)}^{2}y=(xD-\lambda I)z=0.}$$ As before, the solution of ${\displaystyle (xD-\lambda I)z=0}$ is of the form ${\displaystyle z=c_1{x}^{\lambda }}$. So, we are left to solve $${\displaystyle (xD-\lambda I)y=x\frac{dy}{dx}-\lambda y=c_1{x}^{\lambda }.}$$ We then rewrite the equation as $${\displaystyle \frac{dy}{dx}-\frac{\lambda }{x}y=c_1{x}^{\lambda -1},}$$ which one can recognize as being amenable to solution via an integrating factor.

Choose ${\displaystyle M(x)=x^{-\lambda }}$ as our integrating factor. Multiplying our equation through by ${\displaystyle M(x)}$ and recognizing the left-hand side as the derivative of a product, we then obtain $${\displaystyle \begin{array}{cc}\frac{d}{dx}(x^{-\lambda }y)& =c_1{x}^{-1}\\ x^{-\lambda }y& =\int c_1{x}^{-1}dx\\ y& =x^{\lambda }(c_1\ln (x)+c_2)\\ & =c_1\ln (x)x^{\lambda }+c_2{x}^{\lambda }.\end{array}}$$

Example

Given $${\displaystyle x^2{u}^{″}-3x{u}^{\prime }+3u=0,}$$ we substitute the simple solution x^m: $${\displaystyle x^2\left(m\left(m-1\right)x^{m-2}\right)-3x\left(m{x}^{m-1}\right)+3x^m=m\left(m-1\right)x^m-3m{x}^m+3x^m=\left(m^2-4m+3\right)x^m=0.}$$

For x^m to be a solution, either x = 0, which gives the trivial solution, or the coefficient of x^m is zero. Solving the quadratic equation, we get m = 1, 3. The general solution is therefore

${\displaystyle u=c_{1}x+c_2{x}^3.}$