Question

We have seen that $\mathbf{x}\mathbf{=}\mathbf{0}$ is aregular singular point of any Caucher-Euler equation ${\mathbf{ax}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{"}}\mathbf{+}{\mathbf{bxy}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{cy}\mathbf{=}\mathbf{0}$. Are the indicial equation (14) for a Caucher-Euler equation and its auxiliary equation related? Discuss.

Short answer

Yes, it is the same as the auxiliary equation of Cauchy-Euler equation

Step-by-step solution

Definition of Cauchy-Euler equation

A linear differential equation of the form

${\mathbf{a}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}\frac{{\mathbf{d}}^{\mathbf{n}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{n}}}\mathbf{+}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathbf{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\frac{{\mathbf{d}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{n}\mathbf{-}\mathbf{1}}}\mathbf{+}\mathbf{.}\mathbf{..}\mathbf{+}{\mathbf{a}}_{\mathbf{1}}\mathbf{x}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathbf{a}}_{\mathbf{0}}\mathbf{y}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

Where the coefficients${\mathbf{a}}_{\mathbf{n}}\mathbf{,}{\mathbf{a}}_{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{...}\mathbf{,}{\mathbf{a}}_{\mathbf{0}}$are constants, is known as a Cauchy-Euler equation.

We have the form in

${\mathrm{ax}}^2{\mathrm{y}}^{"}+{\mathrm{bxy}}^{\text{'}}+\mathrm{cy}=0$

Which is the Cauchy-Euler equation and has a regular singular point at$x=0$.

Comparing the indicial equation for the given differential equation and its auxiliary equation.

First, without a proof, that if we have a differential equation, with the following standard form;

${\mathrm{y}}^{"}+\mathrm{P}\left(\mathrm{x}\right){\mathrm{y}}^{\text{'}}+\mathrm{Q}\left(\mathrm{x}\right)=0$

Which has a regular singular point at $x=x_0$.

Use Forbinous method to solve differential equation 

The Forbinous method to solve this differential equation is,

$\mathrm{y}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{c}}_{\mathrm{n}}{(\mathrm{x}-{\mathrm{x}}_0)}^{\mathrm{n}+\mathrm{r}}$ …… (1)

We put the given differential equation in the standard form.

$\frac{{\mathrm{ax}}^2}{{\mathrm{ax}}^2}{\mathrm{y}}^{"}+\frac{\mathrm{bx}}{{\mathrm{ax}}^2}{\mathrm{y}}^{\text{'}}+\frac{\mathrm{c}}{{\mathrm{ax}}^2}\mathrm{y}=0$

${\mathrm{y}}^{"}+\frac{\mathrm{b}}{\mathrm{ax}}{\mathrm{y}}^{\text{'}}+\frac{\mathrm{c}}{{\mathrm{ax}}^2}\mathrm{y}=0$

$\mathrm{P}\left(\mathrm{x}\right)=\frac{\mathrm{b}}{\mathrm{ax}}$ and $\mathrm{Q}\left(\mathrm{x}\right)=\frac{\mathrm{c}}{{\mathrm{ax}}^2}$

By using (1) and the Taylor series expansionfor $p\left(x\right)$and $q\left(x\right)$, find the indicial quadratic equation of r.

$r(r-1)+a_{0}r+b_0=0$

Where $a_0$and $b_0$ are the first term of Taylor series of $q\left(x\right)$and $p\left(x\right)$.

$p\left(x\right)=xP\left(x\right)$

$q\left(x\right)=x^{2}Q\left(x\right)$

$\begin{array}{l}=x^2.\frac{c}{a{x}^2}\\ =\frac{c}{a}\end{array}$

Therefore, $a_0=\frac{b}{a}$and $b_0=\frac{c}{a}$

The quadratic equation of r is reduced to the following form:

$r^2-r+\frac{b}{a}r+\frac{c}{a}=0$

Multiplying by $\left(a\right)$, we get

$\left(a\right).r^2-\left(a\right).r+\left(a\right).\frac{b}{a}r+\left(a\right).\frac{c}{a}=0$

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

which is the same as the auxiliary equation ofCauchy-Euler equation.

Therefore, the given equation is the same as the auxiliary equation of Cauchy-Euler equation.