Question

Regular singular points at $\mathbf{x}\mathbf{=}\mathbf{1}$and at$\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}$

Short answer

Answer

$(\mathrm{x}+3{)}^2(\mathrm{x}-1{)}^2{\mathrm{y}}^{"}+(\mathrm{x}+3)(\mathrm{x}-1){\mathrm{y}}^{\text{'}}+\mathrm{y}=0$

Step-by-step solution

To Find a linear second-order differential equation

Consider a linear second-order differential equation in the standard form

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

If the factor $x-1$appears at most to the first power in the denominator of $\mathrm{P}\left(\mathrm{x}\right)$and at most to the second power in the denominator of $\mathrm{Q}\left(\mathrm{x}\right)$, then $\mathrm{x}=1$is a regular singular point. Similarly for $\mathrm{x}=-3$So, let

$\mathrm{P}\left(\mathrm{x}\right)=\frac{1}{\left(\mathrm{x}+3\right)\left(\mathrm{x}-1\right)}\mathrm{and}\mathrm{Q}\left(\mathrm{x}\right)=\frac{1}{{\left(\mathrm{x}+3\right)}^2{\left(\mathrm{x}-1\right)}^2}$

Final Answer

We see that for the above parameters, $\mathrm{x}=1$and $\mathrm{x}=-3$are regular singular points for a linear second-order differential equation. Hence, the equation with the given properties in the standard form is

$\mathrm{y}"+\frac{1}{(\mathrm{X}+3\left(\mathrm{x}-1\right)}\mathrm{y}\text{'}+\frac{1}{\left(\mathrm{x}+3^2\right){(\mathrm{x}-1)}^2}\mathrm{y}=0$

or equivalently

${\left(\mathrm{x}+3\right)}^2+{\left(\mathrm{x}-2\right)}^2\mathrm{y}"+\left(\mathrm{x}+3\right)\left(\mathrm{x}-1\right)\mathrm{y}\text{'}+\mathrm{y}=0$