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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

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

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

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

xy+(1x)yy=0    x2y+(1x)xyxy=0.\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

x2y+pxy+qy=0, where p and q are constants.x^2y''+pxy'+qy=0,\>\>\>\> \text{ where $p$ and $q$ are constants}.

Comparing above two equations we have

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

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

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

Now notice that

y(x)=n=0(s+n)anxs+n1y(x)=n=0(s+n)(s+n1)anxs+n2.\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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