Question

Find all singular points of the given equation and determine whether each one is regular or irregular. \(x^{2}(1-x) y^{\prime \prime}+(x-2) y^{\prime}-3 x y=0\)

Short answer

Question: What are the singular points of the given equation \(x^2(1-x)y''+(x-2)y' -3xy = 0\) and determine if they are regular or irregular. Answer: Both singular points \(x = 0\) and \(x = 1\) are regular singular points.

Step-by-step solution

Identify the equation's singular points

To find the singular points, we first look at the leading coefficient of the highest order derivative, which is \(x^2(1-x)\). A singular point occurs when this leading coefficient is zero. So, we solve the equation: \(x^2(1-x) = 0\) By solving this equation, we obtain the singular points \(x = 0\) and \(x = 1\).

Find the indicial equation

To determine the nature of each singular point, we need to find the corresponding indicial equation. To do that, we can divide the given equation by the leading coefficient of the highest order derivative: \(\frac{x^2(1-x)y''+(x-2)y' -3xy}{x^2(1-x)} = 0\) By simplifying, we get the equation: \(y'' + \frac{(x-2)}{(1-x)}y' - \frac{3}{(1-x)}y = 0\) Now, we can assume a solution in the form of a power series: \(y(x) = \sum_{n=0}^\infty a_n x^{r+n}\), where \(r\) is to be determined. We then compute the derivative (\(y'(x)\)) and the second derivative (\(y''(x)\)): \(y'(x) = \sum_{n=0}^\infty (r+n) a_n x^{r+n-1}\) \(y''(x) = \sum_{n=0}^\infty (r+n)(r+n-1) a_n x^{r+n-2}\) Next, we substitute the series for \(y(x)\), \(y'(x)\), and \(y''(x)\) back into the simplified equation: \(\sum_{n=0}^\infty (r+n)(r+n-1) a_n x^{r+n-2} + \frac{(x-2)}{(1-x)} \sum_{n=0}^\infty (r+n) a_n x^{r+n-1} - \frac{3}{(1-x)} \sum_{n=0}^\infty a_n x^{r+n} = 0\)

Compute indicial equation

To find the indicial equation, we need to evaluate the coefficient of the lowest power of \(x\) for each singular point. For \(x=0\), we look at the coefficient of \(x^{r-2}\) in the first term and \(x^{r-1}\) in the second term. The lowest powers are \(r = 2\) and \(r = 1\) respectively, so the indicial equation must have \(r = 2\). For \(x=1\), we look at the coefficient of \(x^{r+1-2}\) in the first term and \(x^{r-1}\) in the second term. The lowest powers are \(r = 1\) and \(r = 2\) respectively, so the indicial equation must have \(r = 1\).

Determine the nature of singular points

To determine if the singular points are regular or irregular, we must check the root difference of the indicial equation. For \(x = 0\), since \(r = 2\), the root difference is \(2 - 2 = 0\), so the singular point \(x = 0\) is regular. For \(x = 1\), since \(r = 1\), the root difference is \(1 - 1 = 0\), so the singular point \(x = 1\) is regular. In conclusion, both singular points \(x = 0\) and \(x = 1\) are regular singular points.