Euler–Cauchy equation

The general second-order Euler–Cauchy equation is:

$a{x}^2{y}^{″}+bx{y}^{\prime }+cy=0$

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:

$y^{″}+p(x)y^{\prime }+q(x)y=0$

where $p(x)=\frac{b}{ax}$ and $q(x)=\frac{c}{a{x}^2}$.

Then $xp(x)=\frac{b}{a}$ and $x^{2}q(x)=\frac{c}{a}$. 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

$y=\sum _{n=0}^{\mathrm{\infty }}{k}_n{x}^{n+r}$

be a tentative Frobenius-series solution. Then

$y^{\prime }=\sum _{n=0}^{\mathrm{\infty }}{k}_n(n+r)x^{n+r-1}$

$y^{″}=\sum _{n=0}^{\mathrm{\infty }}{k}_n(n+r)(n+r-1)x^{n+r-2}$

Substituting these into the E.C.e. yields

$\sum _{k=0}^{\mathrm{\infty }}{k}_n[a(n+r)(n+r-1)+b(n+r)+c]x^{n+r}=0$

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

$r_1=\frac{-(b-a)+\sqrt{(b-a{)}^2-4ac}}{2a}$

$r_2=\frac{-(b-a)-\sqrt{(b-a{)}^2-4ac}}{2a}$

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.

$\left|\begin{array}{cc}{x}^{r_1}& x^{r_2}\\ r_1{x}^{r_1-1}& r_2{x}^{r_2-1}\end{array}\right|=r_2{x}^{r_1}{x}^{r_2-1}-r_1{x}^{r_1-1}{x}^{r_2}=(r_2-r_1)x^{r_1+r_2-1}$

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 $k_0=\frac{B}{2i}$ and $h_0=\frac{-B}{2i}$ then

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

These two solutions are linearly independent as can be verified by calculating their Wronskian:

$\left|\begin{array}{cc}{x}^{\alpha }\cos (\beta \ln x)& x^{\alpha }\sin (\beta \ln x)\\ x^{\alpha -1}(\alpha \cos (\beta \ln x)-\beta \sin (\beta \ln x))& x^{\alpha -1}(\alpha \sin (\beta \ln x)+\beta \cos (\beta \ln x))\end{array}\right|$

$=x^{2\alpha -1}\beta$

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:

${y_2}^{\prime }=u^{\prime }{x}^r+ur{x}^{r-1}$

${y_2}^{″}=u^{″}{x}^r+2u^{\prime }r{x}^{r-1}+ur(r-1)x^{r-2}$

The original E.C.e. is

$a{x}^2{y}^{″}+bx{y}^{\prime }+cy=0$

Substitute $y_2$ and its derivatives into the E.C.e.:

$a{x}^2(u^{″}{x}^r+2u^{\prime }r{x}^{r-1}+ur(r-1)x^{r-2})+bx(u^{\prime }{x}^r+ur{x}^{r-1})+cu{x}^r=0$

Distribute:

$a{u}^{″}{x}^{r+2}+2a{u}^{\prime }r{x}^{r+1}+aur(r-1)x^r+b{u}^{\prime }{x}^{r+1}+bur{x}^r+cu{x}^r=0$

Group by derivatives of $u$:

$u^{″}a{x}^{r+2}+u^{\prime }(2ar{x}^{r+1}+b{x}^{r+1})+u[ar(r-1)x^r+br{x}^r+c{x}^r]=0$

Factor out powers of $x$:

$u^{″}a{x}^{r+2}+u^{\prime }(2ar+b)x^{r+1}+u[ar(r-1)+br+c]x^r=0$

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.

$u^{″}a{x}^{r+2}+u^{\prime }(2ar+b)x^{r+1}=0$

For $x>0$, $x^{r+1}\ne 0$, so divide by $x^{r+1}$:

$u^{″}ax+u^{\prime }(2ar+b)=0$

And here note well that $2ar+b=a$, thus

$u^{″}ax+u^{\prime }a=0$

$u^{″}x+u^{\prime }=0$

Let $u^{\prime }=v$, so that $v^{\prime }=u^{″}$:

$v^{\prime }x+v=0$

$\frac{v^{\prime }}{v}=-\frac{1}{x}$

$\int \frac{dv}{v}=-\int \frac{dx}{x}$

$\ln |v|=-(\ln |x|+k_0)$

$|v|=e^{-\ln |x|-k_0}=\frac{k_1}{|x|}$

$v=\frac{k_1}{x}=u^{\prime }$

assuming that $x>0$.

$u=\int \frac{k_1}{x}dx=k_1\ln x$

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:

$a{x}^2{y}^{″}+bx{y}^{\prime }+cy=0$

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. $x$?

$\frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}=\frac{1}{x}\frac{dy}{dz}$

$Dy=\frac{1}{x}𝔇y$

where $𝔇=d/dz$.

$xDy=𝔇y$

$𝔇\left(\frac{1}{x}\right)=D\left(\frac{1}{x}\right)𝔇x$

$x=e^z$ so $𝔇x=x$

$𝔇\left(\frac{1}{x}\right)=-\frac{1}{x^2}x=-\frac{1}{x}$

$D^{2}y=\frac{1}{x}𝔇\left(\frac{1}{x}𝔇y\right)$

$D^{2}y=\frac{1}{x}\left[𝔇\right(\frac{1}{x})𝔇y+\frac{1}{x}{𝔇}^{2}y]=\frac{1}{x}[-\frac{1}{x}𝔇y+\frac{1}{x}{𝔇}^{2}y]$

$D^{2}y=\frac{1}{x^2}𝔇(𝔇-1)y$

$x^2{D}^{2}y=𝔇(𝔇-1)y$

So

$(a{x}^2{D}^2+bxD+c)y=0$

becomes

$(a𝔇(𝔇-1)+b𝔇+c)y=0$

$(a{𝔇}^2+(b-a)𝔇+c)y=0$

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

$y=k_1{e}^{rz}+k_{2}z{e}^{rz}$

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

$\frac{d}{dz}\left(\frac{1}{x^2}\right)=\frac{d}{dz}(e^{-2z})=-2e^{-2z}=\frac{-2}{x^2}$

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

$=\frac{1}{x}𝔇\left(\frac{1}{x^2}\right({𝔇}^2-𝔇)y)$

$=\frac{1}{x}\left[𝔇\right(\frac{1}{x^2})({𝔇}^2-𝔇)y+\frac{1}{x^2}𝔇({𝔇}^2-𝔇)y]$

$=\frac{1}{x}[\frac{-2}{x^2}({𝔇}^2-𝔇)y+\frac{1}{x^2}({𝔇}^3-{𝔇}^2)y]$

$=\frac{1}{x^3}({𝔇}^3-3{𝔇}^2+2𝔇)y$

$D^{3}y=\frac{1}{x^3}𝔇(𝔇-1)(𝔇-2)y$

$x^3{D}^{3}y=𝔇(𝔇-1)(𝔇-2)y$

This can be generalized:

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

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:

$D(D^n)y=D[\frac{1}{x^n}𝔇(𝔇-1)...(𝔇-(n-1))y]$

$=\frac{1}{x}𝔇[\frac{1}{x^n}𝔇(𝔇-1)...(𝔇-(n-1))y]$

$=\frac{1}{x}[\frac{-n}{x^n}+\frac{1}{x^n}𝔇]𝔇(𝔇-1)...(𝔇-(n-1))y$

$D^{n+1}y=\frac{1}{x^{n+1}}(𝔇-n)𝔇(𝔇-1)...(𝔇-(n-1))y$

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

$x^{n+1}{D}^{n+1}y=𝔇(𝔇-1)(𝔇-2)...(𝔇-n)y$

Thus the inductive step has been proven; the generalization is true:

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

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 $\ln x$.

External links

• How do I use Frobenious' power series method to solve a Cauchy-Euler's equation ? at Socratic Q&A