# How do I use Frobenious' power series method to solve a Cauchy-Euler's equation ?

Sep 10, 2016

I get to answer my own question! After spending quite some time at the library, I am finally here with how to solve it.

#### Explanation:

Let us consider the standard second order Cauchy-Euler's equation.

$a {x}^{2} \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 + b x \frac{\mathrm{dy}}{\mathrm{dx}} + c y = 0$

For applying the standard method of Frobenious, let $y \left(x\right) = \sum {k}_{n} {x}^{n + l a m \mathrm{da}}$ be the trial solution.

This gives, $\frac{\mathrm{dy}}{\mathrm{dx}} = \sum {k}_{n} \left(n + l a m \mathrm{da}\right) {x}^{n + l a m \mathrm{da} - 1}$

and,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \sum {k}_{n} \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) {x}^{n + l a m \mathrm{da} - 2}$

Substituting these expressions in the original equation,

$a {x}^{2} \sum {k}_{n} \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) {x}^{n + l a m \mathrm{da} - 2} + b x \sum {k}_{n} \left(n + l a m \mathrm{da}\right) {x}^{n + l a m \mathrm{da} - 1} + c \sum {k}_{n} {x}^{n + l a m \mathrm{da}} = 0$

which gives,
$a \sum {k}_{n} \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) {x}^{n + l a m \mathrm{da}} + b \sum {k}_{n} \left(n + l a m \mathrm{da}\right) {x}^{n + l a m \mathrm{da}} + c \sum {k}_{n} {x}^{n + l a m \mathrm{da}} = 0$

further,

$\sum {k}_{n} \left[a \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) + b \left(n + l a m \mathrm{da}\right) + c\right] {x}^{n + l a m \mathrm{da}} = 0$

Now, equating to zero, the coefficient of ${x}^{l} a m \mathrm{da}$ where, $n = 0$,

${k}_{0} \left[a \left(l a m \mathrm{da}\right) \left(l a m \mathrm{da} - 1\right) + b \left(l a m \mathrm{da}\right) + c\right] = 0$

Since ${k}_{0} \ne 0$ $\implies a \left(l a m \mathrm{da}\right) \left(l a m \mathrm{da} - 1\right) + b \left(l a m \mathrm{da}\right) + c = 0$

$\implies a l a m {\mathrm{da}}^{2} - \left(a - b\right) l a m \mathrm{da} + c = 0$

This is the indical equation. Solving for $l a m \mathrm{da}$,

$l a m {\mathrm{da}}_{1} = \frac{\left(a - b\right) + \sqrt{{\left(a - b\right)}^{2} - 4 a c}}{2 a}$ ; $l a m {\mathrm{da}}_{2} = \frac{\left(a - b\right) - \sqrt{{\left(a - b\right)}^{2} - 4 a c}}{2 a}$

Where $l a m {\mathrm{da}}_{1}$ and $l a m {\mathrm{da}}_{2}$ are the two roots of the indical equation.

considering now,

$\sum {k}_{n} \left[a \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) + b \left(n + l a m \mathrm{da}\right) + c\right] {x}^{n + l a m \mathrm{da}} = 0$

now, equating to zero the coefficient of ${x}^{n + l a m \mathrm{da}}$,

${k}_{n} \left[a \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) + b \left(n + l a m \mathrm{da}\right) + c\right] = 0$

Now for each value of $l a m \mathrm{da}$ substituted in the above equation, we shall obtain that ${k}_{0}$ and some other one coefficient is non zero while all others are zero.

Thus, for each of the two $l a m \mathrm{da}$, we shall obtain one solution.

If ${y}_{1}$ and ${y}_{2}$ are the two solutions (each containing one or two terms each), then the general solution is obtained by superposition of the two solutions (this is possible since, the equation is linear).

Thus, $y = A {y}_{1} + B {y}_{2}$ where A and B are arbitrary constants.

As an example, let us consider a very simple version of the equation,

${x}^{2} \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 - 6 y = 0$ where, $a = 1 , b = 0$ and $c = - 6$

I've taken this equation, just for the purpose of illustrating the process. You can actually do it with any such equation with $b \ne 0$ ofcourse.

So, let, $y \left(x\right) = \sum {k}_{n} {x}^{n + l a m \mathrm{da}}$ be the trial solution.

This gives, $\frac{\mathrm{dy}}{\mathrm{dx}} = \sum {k}_{n} \left(n + l a m \mathrm{da}\right) {x}^{n + l a m \mathrm{da} - 1}$

and,

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \sum {k}_{n} \left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) {x}^{n + l a m \mathrm{da} - 2}$

Substituting these in the original equation, and doing simplifying,

$\sum {k}_{n} \left[\left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) - 6\right] {x}^{n + l a m \mathrm{da}} = 0$

Now equating to zero, the coefficient of ${x}^{n + l a m \mathrm{da}}$

${k}_{0} \left[\left(l a m \mathrm{da}\right) \left(l a m \mathrm{da} - 1\right) - 6\right] = 0$ and since ${k}_{0} \ne 0$

implies, $\left[\left(l a m \mathrm{da}\right) \left(l a m \mathrm{da} - 1\right) - 6\right] = 0$

which on simplifying gives, $l a m {\mathrm{da}}_{1} = - 2$ and $l a m {\mathrm{da}}_{2} = 3$

Consider now, $\sum {k}_{n} \left[\left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) - 6\right] {x}^{n + l a m \mathrm{da}} = 0$

Equating to zero, the coefficient of ${x}^{n + l a m \mathrm{da}}$
${k}_{n} \left[\left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) - 6\right] = 0$

Assuming ${k}_{n} \ne 0$ , $\left[\left(n + l a m \mathrm{da}\right) \left(n + l a m \mathrm{da} - 1\right) - 6\right] = 0$

For $l a m \mathrm{da} = - 2$

$\left[\left(n - 2\right) \left(n - 3\right) - 6\right] = 0$
$\implies {n}^{2} - 5 n = 0$

which gives, $n = 0 , 5$ thus, for $l a m {\mathrm{da}}_{1} = - 2$, ${k}_{0} , {k}_{5} \ne 0$

For $l a m \mathrm{da} = 3$

$\left[\left(n + 3\right) \left(n + 2\right) - 6\right] = 0$
$\implies {n}^{2} + 5 n = 0$

Thus, $n = 0 , - 5$

Thus, only ${k}_{0} , {k}_{-} 5 \ne 0$. Discarding ${k}_{-} 5$ for $l a m {\mathrm{da}}_{2} = 3 , {k}_{0} \ne 0$

Thus, if ${y}_{1}$ is solution corresponding to $l a m {\mathrm{da}}_{1} = - 2$

${y}_{1} = \sum {k}_{n} {x}^{n - 2} \implies {y}_{1} = {k}_{0} {x}^{-} 2 + {k}_{5} {x}^{5 - 2}$

which gives, ${y}_{1} = {k}_{0} {x}^{-} 2 + {k}_{5} {x}^{3}$ (All other terms are zero because the coefficients are zero other than ${k}_{0} \mathmr{and} {k}_{5}$)

Similarly, for $l a m {\mathrm{da}}_{2} = 3$; ${y}_{2} = \sum {k}_{n} {x}^{n + 3}$

Implies, ${y}_{2} = {k}_{0} {x}^{3}$ (All other terms are zero other than ${k}_{0}$)

Thus, the general solution, $y = A {y}_{1} + B {y}_{2}$

$\implies y = A {k}_{0} {x}^{-} 2 + A {k}_{5} {x}^{3} + B {k}_{0} {x}^{3}$

putting ${C}_{1} = A {k}_{0}$ and ${C}_{2} = A {k}_{5} + B {k}_{0}$

Thus, we have $y = {C}_{1} {x}^{-} 2 + {C}_{2} {x}^{3}$ where ${C}_{1}$ and ${C}_{2}$ are completely arbitrary constants.