## Related questions with answers

For the differential equations, find the indicial polynomial for the singularity at x=0. Then find the recurrence formula for the largest of the roots to the indicial equation. xy''+(1-x)y'-y=0

Solution

Verified$\textbf{Solution}$: Let us consider the given differential equation

$\begin{align}xy''+(1-x)y'-y=0.\end{align}$

Now notice that, if we recasted the above equation we have

$\begin{align*} xy''+(1-x)y'-y=0 \implies \> &x^2y''+(1-x)xy'-xy=0.\end{align*}$

Let us now consider the general Euler's equation as

$x^2y''+pxy'+qy=0,\>\>\>\> \text{ where $p$ and $q$ are constants}.$

Comparing above two equations we have

$p=1-x\>\>\> \text{ and }\>\>\> q=-x.$

From the aforesaid argument we have, both $p$ and $q$ are analytic at $x=0$. So, $x=0$ is a regular singular point. Let us now look at the Frobenius solution as

$y(x)=\sum_{n=0}^{\infty} a_n x^{s+n}.$

Now notice that

$\begin{align*} y'(x) &= \sum_{n=0}^{\infty} (s+n) a_n x^{s+n-1} \\ y''(x) &= \sum_{n=0}^{\infty} (s+n)(s+n-1) a_n x^{s+n-2}. \end{align*}$

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