Euler–Cauchy equation

The general second-order Euler–Cauchy equation is:
 * $$a x^2 y'' + b x y' + c y = 0 $$

where a, b, and c can be any (real) constants.

Solving by the Method of Frobenius
The leading coefficient function $$a x^2$$ 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 $$p(x) = {b\over a x}$$ and $$q(x) = {c\over a x^2}$$.

Then $$x p(x) = {b \over a}$$ and $$x^2 q(x) = {c \over a}$$. These two functions, $$x p(x)$$ 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 $$n=0$$,
 * $$ k_0 (a r^2 + (b-a)r + c) = 0 $$

Letting $$k_0 \ne 0$$, then
 * $$ a r^2 + (b - a)r + c = 0 $$

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 $$n\ne 0$$ and $$n\ne n_1$$. (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 $$r_2 \ne r_1$$ 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
 * $$ y = A x^\alpha \cos(\beta \ln x)$$

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
 * $$ y = B x^\alpha \sin (\beta \ln x)$$

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
 * $$ y = A x^\alpha \cos(\beta \ln x) + B x^\alpha \sin(\beta \ln x)$$

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
 * $$ y_2 = u x^r $$

where $$u$$ is a function of $$x$$. Now derive:
 * $$ y_2' = u' x^r + u r x^{r-1} $$


 * $$ y_2 = u x^r + 2 u' r x^{r-1} + u r (r-1) x^{r-2} $$

The original E.C.e. is
 * $$ a x^2 y'' + b x y' + c y = 0 $$

Substitute $$y_2$$ and its derivatives into the E.C.e.:
 * $$ a x^2 (u'' x^r + 2 u' r x^{r-1} + ur(r-1)x^{r-2}) + bx(u' x^r + u r x^{r-1}) + c u x^r = 0 $$

Distribute:
 * $$ a u'' x^{r+2} + 2 a u' r x^{r+1} + a u r (r-1) x^r + b u' x^{r+1} + b u r x^r + c u x^r = 0 $$

Group by derivatives of $$u$$:
 * $$ u'' a x^{r+2} + u' (2 a r x^{r+1} + b x^{r+1}) + u [a r (r-1) x^r + b r x^r + c x^r] = 0 $$

Factor out powers of $$x$$:
 * $$ u'' a x^{r+2} + u' (2 a r + b) x^{r+1} + u [a r (r-1) + b r + c] x^r = 0 $$

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.
 * $$ u'' a x^{r+2} + u' (2 a r + b) x^{r+1} = 0 $$

For $$x>0$$, $$x^{r+1} \ne 0$$, so divide by $$x^{r+1}$$:
 * $$ u''a x + u' (2 a r + b) = 0 $$

And here note well that $$2 a r + b = a$$, thus
 * $$ u'' a x + u' a = 0 $$


 * $$ u'' x + u' = 0 $$

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 $$y_2 = u x^r = k_1 x^r \ln x$$ 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:
 * $$ a x^2 y'' + b x y' + c y = 0 $$

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. $$x$$?


 * $${dy \over dx} = {dy \over dz} {dz \over dx} = {1\over x} {dy \over dz} $$


 * $$ D y = {1\over x} \mathfrak{D} y $$

where $$\mathfrak{D} = d / dz$$.

$$ x D y = \mathfrak{D} y $$


 * $$ \mathfrak{D}\Big({1\over x}\Big) = D\Big({1\over x}\Big) \mathfrak{D} x$$

$$x = e^z$$ so $$\mathfrak{D} x = x$$


 * $$ \mathfrak{D}\Big({1\over x}\Big) = -{1\over x^2} x = -{1\over x} $$


 * $$ D^2 y = {1\over x} \mathfrak{D} \Big({1\over x} \mathfrak{D} y\Big) $$


 * $$ D^2 y = {1\over x} \Bigg[\mathfrak{D} \Big({1\over x}\Big) \mathfrak{D} y + {1\over x} \mathfrak{D}^2 y\Bigg] = {1\over x} \Bigg[-{1\over x}\mathfrak{D} y + {1\over x} \mathfrak{D}^2 y\Bigg] $$


 * $$ D^2 y = {1\over x^2} \mathfrak{D} (\mathfrak{D} - 1) y $$

$$ x^2 D^2 y = \mathfrak{D} (\mathfrak{D} - 1) y $$

So
 * $$ (a x^2 D^2 + b x D + c) y = 0 $$

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
 * $$ a r^2 + (b - a) r + c = 0 $$

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 $$ y = k_1 x^{r_1} + k_2 x^{r_2} $$

If the characteristic equation has two equal (and thus real) roots then the H.O.L.D.E.w.c.c. has solution
 * $$ y = k_1 e^{r z} + k_2 z e^{r z} $$

so the E.C.e. has solution $$ y = k_1 x^r + k_2 x^r \ln x$$

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 $$ y = k_1 x^\alpha \cos(\beta \ln x) + k_2 x^\alpha \sin(\beta \ln x) $$

Higher-order cases

 * $$ {d\over dz} \Big({1\over x^2}\Big) = {d\over dz} (e^{-2z}) = -2 e^{-2z} = {-2 \over x^2} $$


 * $$ D^3 y = D D^2 y = D \Big({1\over x^2} \mathfrak{D} (\mathfrak{D} - 1) y\Big) $$


 * $$ \qquad = {1\over x} \mathfrak{D} \Bigg({1\over x^2} \Big(\mathfrak{D}^2 - \mathfrak{D}) y\Bigg) $$


 * $$ \qquad = {1\over x} \Bigg[\mathfrak{D}\Big({1\over x^2}\Big) (\mathfrak{D}^2 - \mathfrak{D}) y + {1\over x^2} \mathfrak{D} (\mathfrak{D}^2 - \mathfrak{D}) y\Bigg] $$


 * $$ \qquad = {1\over x} \Bigg[{-2\over x^2} (\mathfrak{D}^2 - \mathfrak{D}) y + {1\over x^2} (\mathfrak{D}^3 - \mathfrak{D}^2) y\Bigg] $$


 * $$ \qquad = {1\over x^3} (\mathfrak{D}^3 - 3 \mathfrak{D}^2 + 2 \mathfrak{D}) y $$


 * $$ \qquad D^3 y = {1\over x^3} \mathfrak{D} (\mathfrak{D} - 1) (\mathfrak{D} - 2) y $$


 * $$ x^3 D^3 y = \mathfrak{D} (\mathfrak{D} - 1) (\mathfrak{D} - 2) y $$

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: $$ x^n D^n y = \mathfrak{D} (\mathfrak{D} - 1) (\mathfrak{D} - 2) ... (\mathfrak{D} - (n - 1)) y $$

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 $$\ln x$$.