# Euler–Cauchy equation

The general second-order Euler–Cauchy equation is:
    $ax^{2}y''+bxy'+cy=0$
where a, b, and c can be any (real) constants.

## Solving by the Method of Frobenius

The leading coefficient function $ax^{2}$ is 0 at $x=0$; so there is a singular point at $x=0$. Normal form is:
    $y''+p(x)y'+q(x)y=0$
where $p(x)={b \over ax}$ and $q(x)={c \over ax^{2}}$.
Then $xp(x)={b \over a}$ and $x^{2}q(x)={c \over a}$. These two functions, $xp(x)$ and $x^{2}q(x)$, have no discontinuities; so the singular point is regular. Thus the method of Frobenius is applicable to this E.C.e. at $x=0$.
Let
    $y=\sum _{n=0}^{\infty }k_{n}x^{n+r}$
be a tentative Frobenius-series solution. Then
    $y'=\sum _{n=0}^{\infty }k_{n}(n+r)x^{n+r-1}$
    $y''=\sum _{n=0}^{\infty }k_{n}(n+r)(n+r-1)x^{n+r-2}$
Substituting these into the E.C.e. yields
    $\sum _{k=0}^{\infty }k_{n}[a(n+r)(n+r-1)+b(n+r)+c]x^{n+r}=0$
Since the L.H.S. is identically equal to 0, then each coefficient of the Frobenius series must equal 0:
    $k_{n}[a(n+r)(n+r-1)+b(n+r)+c]=0$
    $k_{n}[a(n+r)^{2}+(b-a)(n+r)+c]=0$
When $n=0$,
    $k_{0}(ar^{2}+(b-a)r+c)=0$
Letting $k_{0}\neq 0$, then
    $ar^{2}+(b-a)r+c=0$
is the indicial equation. Its roots are
    $r_{1}={-(b-a)+{\sqrt {(b-a)^{2}-4ac}} \over 2a}$
    $r_{2}={-(b-a)-{\sqrt {(b-a)^{2}-4ac}} \over 2a}$
If there exists some integer $n_{1}$ such that
    $a(n_{1}+r)^{2}+(b-a)(n_{1}+r)+c=0$
then either $r_{2}=n_{1}+r_{1}$ or $r_{1}=n_{1}+r_{2}$ (since $r_{1}$ and $r_{2}$ are the only two solutions of the indicial equation); and there can be at most one such $n_{1}$. For all other values of $n$, the equation
    $a(n+r)^{2}+(b-a)(n+r)+c=0$
is not satisfied, which means that $k_{n}=0$ for all $n$ such that $n\neq 0$ and $n\neq n_{1}$. (N.B.: If $r_{2}-r_{1}$ is not an integer then such an $n_{1}$ does not exist.)
    $y_{1}=k_{0}x^{r_{1}}$ or $y_{1}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{1}+n_{1}}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{2}}$
    $y_{2}=h_{0}x^{r_{2}}$ or $y_{2}=h_{0}x^{r_{2}}+h_{n_{1}}x^{r_{2}+n_{1}}=h_{0}x^{r_{2}}+h_{n_{1}}x^{r_{1}}$
The two right sides of the two logical disjunctions cannot be simultaneously true. Since $k_{n_{1}}$ is independent of $k_{0}$, then a solution such as $y_{1}=k_{0}x^{r_{1}}+k_{n_{1}}x^{r_{2}}$ is a linear combination of linearly independent solutions.
    ${\Bigg |}{\begin{matrix}x^{r_{1}}&x^{r_{2}}\\r_{1}x^{r_{1}-1}&r_{2}x^{r_{2}-1}\end{matrix}}{\Bigg |}=r_{2}x^{r_{1}}x^{r_{2}-1}-r_{1}x^{r_{1}-1}x^{r_{2}}=(r_{2}-r_{1})x^{r_{1}+r_{2}-1}$
Assuming that $r_{2}\neq r_{1}$ then this Wronskian is never zero, so the two terms of the solution $y_{1}$ are actually linearly independent solutions. (The case when $r_{2}=r_{1}$ will be dealt with further down from here.) The second term of $y_{1}$ is just “echoing” the other solution $y_{2}=h_{0}x^{r_{2}}$, so to speak. So the general solution is
    $y=k_{0}x^{r_{1}}+h_{0}x^{r_{2}}$

### The case of complex conjugate indicial roots

If $r_{1,2}$ are a complex conjugate pair,
    $r_{1,2}=\alpha \pm i\beta$
then
    $y=k_{0}x^{\alpha }x^{i\beta }+h_{0}x^{\alpha }x^{-i\beta }$
    $y=k_{0}x^{\alpha }e^{i\beta \ln x}+h_{0}x^{\alpha }e^{-i\beta \ln x}$
If $k_{0}$ is set to $A/2$ and $h_{0}$ is set to $A/2$ as well then
    $y=Ax^{\alpha }\cos(\beta \ln x)$
as may be derived through Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$. If $k_{0}={B \over 2i}$ and $h_{0}={-B \over 2i}$ then
    $y=Bx^{\alpha }\sin(\beta \ln x)$
These two solutions are linearly independent as can be verified by calculating their Wronskian:
    ${\Bigg |}{\begin{matrix}x^{\alpha }\cos(\beta \ln x)&x^{\alpha }\sin(\beta \ln x)\\x^{\alpha -1}(\alpha \cos(\beta \ln x)-\beta \sin(\beta \ln x))&x^{\alpha -1}(\alpha \sin(\beta \ln x)+\beta \cos(\beta \ln x))\end{matrix}}{\Bigg |}$
    $=x^{2\alpha -1}\beta$
which is non-zero almost everywhere. Thus
    $y=Ax^{\alpha }\cos(\beta \ln x)+Bx^{\alpha }\sin(\beta \ln x)$
is a general solution.

### The case of equal indicial roots

If $r_{1}=r_{2}$, let $r=r_{1}=r_{2}$;
    $y_{1}=x^{r}$
is one solution. The other one may be determined through the method of undetermined coefficients. Let
    $y_{2}=ux^{r}$
where $u$ is a function of $x$. Now derive:
    $y_{2}'=u'x^{r}+urx^{r-1}$
    $y_{2}''=u''x^{r}+2u'rx^{r-1}+ur(r-1)x^{r-2}$
The original E.C.e. is
    $ax^{2}y''+bxy'+cy=0$
Substitute $y_{2}$ and its derivatives into the E.C.e.:
    $ax^{2}(u''x^{r}+2u'rx^{r-1}+ur(r-1)x^{r-2})+bx(u'x^{r}+urx^{r-1})+cux^{r}=0$
Distribute:
    $au''x^{r+2}+2au'rx^{r+1}+aur(r-1)x^{r}+bu'x^{r+1}+burx^{r}+cux^{r}=0$
Group by derivatives of $u$:
    $u''ax^{r+2}+u'(2arx^{r+1}+bx^{r+1})+u[ar(r-1)x^{r}+brx^{r}+cx^{r}]=0$
Factor out powers of $x$:
    $u''ax^{r+2}+u'(2ar+b)x^{r+1}+u[ar(r-1)+br+c]x^{r}=0$
The expression within the square brackets is the indicial equation, for which $r$ is the root, so the third term in the sum of the L.H.S. vanishes.
    $u''ax^{r+2}+u'(2ar+b)x^{r+1}=0$
For $x>0$, $x^{r+1}\neq 0$, so divide by $x^{r+1}$:
    $u''ax+u'(2ar+b)=0$
And here note well that $2ar+b=a$, thus
    $u''ax+u'a=0$
    $u''x+u'=0$
Let $u'=v$, so that $v'=u''$:
    $v'x+v=0$
    ${v' \over v}=-{1 \over x}$
    $\int {dv \over v}=-\int {dx \over x}$
    $\ln |v|=-(\ln |x|+k_{0})$
    $|v|=e^{-\ln |x|-k_{0}}={k_{1} \over |x|}$
    $v={k_{1} \over x}=u'$
assuming that $x>0$.
    $u=\int {k_{1} \over x}dx=k_{1}\ln x$
So $y_{2}=ux^{r}=k_{1}x^{r}\ln x$ and the general solution is
    $y=k_{0}x^{r}+k_{1}x^{r}\ln x$

## Solving another way

Recalling the second order Euler–Cauchy equation:
    $ax^{2}y''+bxy'+cy=0$
Let $z=\ln x$. This is suggested by the solution $x^{\alpha }\cos(\beta \ln x)$ corresponding to the indicial root $\alpha +i\beta$, with $\cos(\beta \ln x)$ instead of $\cos(\beta x)$ and $\sin(\beta \ln x)$ instead of $\sin(\beta x)$. So, letting $z=\ln x$, what happens when $y$ is derived with respect to $z$ instead of w.r.t. $x$?
    ${dy \over dx}={dy \over dz}{dz \over dx}={1 \over x}{dy \over dz}$
    $Dy={1 \over x}{\mathfrak {D}}y$
where ${\mathfrak {D}}=d/dz$.
    ${\mathfrak {D}}{\Big{1 \over x}{\Big )}=D{\Big{1 \over x}{\Big )}{\mathfrak {D}}x$
$x=e^{z}$ so ${\mathfrak {D}}x=x$
    ${\mathfrak {D}}{\Big{1 \over x}{\Big )}=-{1 \over x^{2}}x=-{1 \over x}$
    $D^{2}y={1 \over x}{\mathfrak {D}}{\Big{1 \over x}{\mathfrak {D}}y{\Big )}$
    $D^{2}y={1 \over x}{\Bigg{\mathfrak {D}}{\Big{1 \over x}{\Big )}{\mathfrak {D}}y+{1 \over x}{\mathfrak {D}}^{2}y{\Bigg ]}={1 \over x}{\Bigg-{1 \over x}{\mathfrak {D}}y+{1 \over x}{\mathfrak {D}}^{2}y{\Bigg ]}$
    $D^{2}y={1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}-1)y$
So
    $(ax^{2}D^{2}+bxD+c)y=0$
becomes
    $(a{\mathfrak {D}}({\mathfrak {D}}-1)+b{\mathfrak {D}}+c)y=0$
    $(a{\mathfrak {D}}^{2}+(b-a){\mathfrak {D}}+c)y=0$
which is a “H.O.L.D.E.” (Homogeneous Ordinary Linear Differential Equation) with constant coefficients and characteristic equation
    $ar^{2}+(b-a)r+c=0$
If its two roots $r_{1,2}$ are real and distinct then the H.O.L.D.E.w.c.c. has solution
    $y=k_{1}e^{r_{1}z}+k_{2}e^{r_{2}z}$
where $z=\ln x$ so the E.C.e. has solution
If the characteristic equation has two equal (and thus real) roots then the H.O.L.D.E.w.c.c. has solution
    $y=k_{1}e^{rz}+k_{2}ze^{rz}$
so the E.C.e. has solution
If the roots are a complex conjugate pair then the H.O.L.D.E.w.c.c. has solution
    $y=k_{1}e^{\alpha z}\cos(\beta z)+k_{2}e^{\alpha z}\sin(\beta z)$
where, again, $z=\ln x$ so the E.C.e. has solution

### Higher-order cases

    ${d \over dz}{\Big{1 \over x^{2}}{\Big )}={d \over dz}(e^{-2z})=-2e^{-2z}={-2 \over x^{2}}$
    $D^{3}y=DD^{2}y=D{\Big{1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}-1)y{\Big )}$
    $\qquad ={1 \over x}{\mathfrak {D}}{\Bigg{1 \over x^{2}}{\Big{\mathfrak {D}}^{2}-{\mathfrak {D}})y{\Bigg )}$
    $\qquad ={1 \over x}{\Bigg{\mathfrak {D}}{\Big{1 \over x^{2}}{\Big )}({\mathfrak {D}}^{2}-{\mathfrak {D}})y+{1 \over x^{2}}{\mathfrak {D}}({\mathfrak {D}}^{2}-{\mathfrak {D}})y{\Bigg ]}$
    $\qquad ={1 \over x}{\Bigg{-2 \over x^{2}}({\mathfrak {D}}^{2}-{\mathfrak {D}})y+{1 \over x^{2}}({\mathfrak {D}}^{3}-{\mathfrak {D}}^{2})y{\Bigg ]}$
    $\qquad ={1 \over x^{3}}({\mathfrak {D}}^{3}-3{\mathfrak {D}}^{2}+2{\mathfrak {D}})y$
    $\qquad D^{3}y={1 \over x^{3}}{\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)y$
    $x^{3}D^{3}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)y$
This can be generalized:
    $x^{n}D^{n}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-(n-1))y$
To prove it, use mathematical induction. The base cases have already been proven. As inductive hypothesis, take this very formula that is to be proven. Divide it by $x^{n}$ and apply $D$ to (both sides of) it:
    $D(D^{n})y=D{\Bigg{1 \over x^{n}}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y{\Bigg ]}$
    $\qquad ={1 \over x}{\mathfrak {D}}{\Bigg{1 \over x^{n}}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y{\Bigg ]}$
    $\qquad ={1 \over x}{\Bigg{-n \over x^{n}}+{1 \over x^{n}}{\mathfrak {D}}{\Bigg ]}{\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y$
    $D^{n+1}y={1 \over x^{n+1}}({\mathfrak {D}}-n){\mathfrak {D}}({\mathfrak {D}}-1)...({\mathfrak {D}}-(n-1))y$
    $D^{n+1}y={1 \over x^{n+1}}{\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-n)y$
    $x^{n+1}D^{n+1}y={\mathfrak {D}}({\mathfrak {D}}-1)({\mathfrak {D}}-2)...({\mathfrak {D}}-n)y$
Thus the inductive step has been proven; the generalization is true:
This rule can be used to convert a higher-order E.C.e. into a H.O.L.D.E. with constant coefficients whose independent variable is z. Once the H.O.L.D.E.w.c.c. is solved, replace z with $\ln x$.