Question

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x^{2}\left(1-x^{2}\right) y^{\prime \prime}+(2 / x) y^{\prime}+4 y=0\)

Short answer

Answer: The singular points of the given equation are \(x = 0, \pm 1\). The singular points \(x= \pm 1\) are regular, while the singular point \(x=0\) is irregular.

Step-by-step solution

Identify Singular Points

First, we need to find the singular points of the given equation by identifying the points where the coefficients of \(y''\) and \(y'\) become undefined or infinite. For the coefficient of \(y''\), we have the term \(x^2(1-x^2)\). This term becomes undefined when \(x^2=0\) or \(x^2=1\), meaning the singular points are \(x=0\) and \(x=\pm1\). For the coefficient of \(y'\), we have the term \((2/x)\). This term becomes undefined when \(x=0\), which is already a singular point from our previous analysis. So, the singular points of the given equation are \(x = 0, \pm 1\).

Determine Regular or Irregular Singular Points

Now, we need to check whether these singular points are regular or irregular. For a singular point to be regular, the coefficients of lower-order derivatives must have converging power series around the singular point. If not, the singular point is irregular. For \(x=\pm1\), we can write the coefficients of the derivatives in the following form: - Coefficient of \(y''\): \((1-x^2) = (1-1)^2 = 0\) and \((1-(-1)^2) = 0\). These coefficients vanish at the singular points and therefore also converge. - Coefficient of \(y'\): \((2/x) = 2\) for \(x = 1\), and \(-2\) for \(x = -1\). Both of these are finite and converging around their respective singular points. Therefore, the singular points \(x= \pm 1\) are regular singular points. Now, let's examine the singular point at \(x=0\): - Coefficient of \(y''\): \((1-x^2) = 1\). This coefficient does not vanish at the singular point but is finite and converging. - Coefficient of \(y'\): \((2/x)\). This term becomes infinite as \(x\) approaches zero. We can find its limit: \(\lim_{x\to 0}\frac{2}{x}=\infty\) Since the coefficient of \(y'\) becomes infinite at \(x=0\), the singular point at \(x=0\) is irregular. So, the given equation has one irregular singular point, \(x=0\), and regular singular points at \(x=\pm 1\).