Question

A regular singular point at \(x = 1\)and an irregular singular point at \(x = 0\)

Short answer

Answer

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

Step-by-step solution

To Find the first three non-zero terms of a power series

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

\(P(x){y^\prime } + Q(x)y = 0\)

If the factor\(x - 1\)appears at most to the first power in the denominator of\(P(x)\)and at most to the second power in the denominator of\(Q(x)\), then\(x = 1\)is a regular singular point. On the contrary, for\(x = 0\)to be a irregular singular point, the factor\(x - 0\)or\(x\)need to appear at least to the second power in the denominator of\(P(x)\)or at least to the third power in the denominator of\(Q(x)\). So, let

\(P(x) = \frac{1}{{{x^2}(x - 1)}}\;\;\; and \;\;\;Q(x) = \frac{1}{{{x^2}{{(x - 1)}^2}}}\)

Final Answer

We see that for the above parameters,\(x = 1\)and\(x = 0\)are regular and irregular singular points for a linear second-order differential equation. Hence, the equation with the given properties in the standard form is

\({y''} + \frac{1}{{{x^2}(x - 1)}}{y'} + \frac{1}{{{x^2}{{(x - 1)}^2}}}y = 0\)

or equivalently

\({x^2}{(x - 1)^2}{y} + (x - 1){y'} + y = 0\)