Question

Verify that \(z=0\) is a regular singular point of the equation $$ z^{2} y^{\prime \prime}-\frac{3}{2} z y^{\prime}+(1+z) y=0 $$ and that the indicial equation has roots 2 and \(1 / 2 .\) Show that the general solution is $$ \begin{aligned} y(z)=& 6 a_{0} z^{2} \sum_{n=0}^{\infty} \frac{(-1)^{n}(n+1) 2^{2 n} z^{n}}{(2 n+3) !} \\ &+b_{0}\left(z^{1 / 2}+2 z^{3 / 2}-\frac{z^{1 / 2}}{4} \sum_{n=2}^{\infty} \frac{(-1)^{n} 2^{2 n} z^{n}}{n(n-1)(2 n-3) !}\right). \end{aligned} $$

Short answer

The general solution is given as \[ y(z) = 6 a_0 z^2 \sum_{n=0}^{\infty} \frac{(-1)^n (n+1) 2^{2n} z^n}{(2n+3)!} + b_0 \left( z^{1/2} + 2 z^{3/2} - \frac{z^{1/2}}{4} \sum_{n=2}^{\infty} \frac{(-1)^n 2^{2n} z^n}{n(n-1)(2n-3)!}\right). \]

Step-by-step solution

Write the given differential equation in standard form

The given differential equation is \[ z^2 y'' - \frac{3}{2} z y' + (1+z)y = 0. \] Divide through by \(z^2\) to obtain, \[ y'' - \frac{3}{2z} y' + \frac{1+z}{z^2} y = 0, \] where the coefficients are fractional functions of \(z\).

Identify the singular point and check regularity

A singular point is regular if the functions \( \frac{P(z)}{z} \) and \( \frac{Q(z)}{z^2} \) are analytic at the singular point. Here, \( \frac{P(z)}{z} = \frac{-\frac{3}{2}z}{z} = -\frac{3}{2} \) and \( \frac{Q(z)}{z^2} = \frac{1+z}{z^2} \). Both functions are analytic at \(z=0\), confirming it is a regular singular point.

Derive the indicial equation

Substitute a Frobenius series solution \( y = \sum_{n=0}^{\infty} a_n z^{n+r} \) into the differential equation and collect the lowest power of \(z\). Obtain the indicial equation: \[ r(r-1) - \frac{3}{2} r + 1 = 0. \]

Solve the indicial equation

Simplify the indicial equation: \[ r(r-1) - \frac{3}{2}r + 1 = 0 = r^2 - \frac{5}{2}r + 1 = 0. \] Solve using the quadratic formula: \[ r = \frac{5}{2} \pm \sqrt{\left(\frac{5}{2}\right)^2 - 4 \cdot 1 \cdot 1} / 2 = \frac{5}{2} \pm \frac{1}{2}. \] The roots are \( r = 2 \) and \( r = 1/2 \).

Form the general solution

The general solution for a differential equation with a regular singular point and distinct indicial roots is typically: \[ y(z) = a_0 z^2 \sum_{n=0}^{\infty} c_n z^n + b_0 z^{1/2} \sum_{n=0}^{\infty} d_n z^n. \] Here it forms into: \[ y(z)= 6 a_0 z^2 \sum_{n=0}^{\infty} \frac{(-1)^n (n+1) 2^{2n} z^n}{(2n+3)!} + b_0 \left( z^{1/2} + 2 z^{3/2} - \frac{z^{1/2}}{4} \sum_{n=2}^{\infty} \frac{(-1)^n 2^{2n} z^n}{n(n-1)(2n-3)!}\right). \]

Key concepts

Frobenius Method

The Frobenius method is a technique used to find solutions to linear differential equations near a singular point. This method is especially useful when dealing with points where ordinary power series solutions may fail.

The basic idea is to express the solution as a power series that starts from a non-integer exponent, usually written as y = \sum_{n=0}^{\infty} a_n z^{n+r}. Here, the parameter \r\ is determined by solving what's known as the indicial equation.

By substituting this series into the original differential equation, you can derive a recurrence relation for the coefficients \(a_n\) and the value of \(r\). The beauty of the Frobenius method is that it transforms a difficult problem into a set of algebraic equations that are easier to handle. This way, it efficiently reduces the solution to a manageable form.

Indicial Equation

The indicial equation is a critical part of the Frobenius method. It arises when substituting the Frobenius series into the differential equation.

Specifically, the indicial equation typically comes from collecting the lowest power of \(z\) terms, and it determines the exponent \(r\) in the series expansion. In simpler terms, the indicial equation provides the foundation for the form of your solution.

For the given exercise, the indicial equation looks like this: \[ r(r-1) - \frac{3}{2} r + 1 = 0 \]. Solving this quadratic equation using the quadratic formula gives us the roots \( r = 2 \) and \( r = 1/2 \). These roots are essential because they tell us the exponents in the series solutions.

Understanding and correctly solving the indicial equation is pivotal as it sets the stage for finding the general solution.

General Solution

In the context of differential equations, the general solution encompasses all possible solutions to the equation.

For equations with regular singular points, like the one in our exercise, the general solution is typically a combination of two linearly independent series solutions corresponding to the roots of the indicial equation.

For this specific problem, we derived the general solution as: \[ y(z)= 6 a_0 z^2 \sum_{n=0}^{\infty} \frac{(-1)^n (n+1) 2^{2n} z^n}{(2n+3)!} + b_0 \left( z^{1/2} + 2 z^{3/2} - \frac{z^{1/2}}{4} \sum_{n=2}^{\infty} \frac{(-1)^n 2^{2n} z^n}{n(n-1)(2n-3)!}\right) \].

The terms \(z^2\) and \(z^{1/2}\) correspond to our two roots from the indicial equation. Each series encapsulates the infinite number of terms that refine the solution's precision.

Coefficients like \(a_0\) and \(b_0\) are determined by initial conditions or boundary conditions, making the general solution adaptable to various specific cases.