Question

Find the general power series solution about \(z=0\) of the equation $$ z \frac{d^{2} y}{d z^{2}}+(2 z-3) \frac{d y}{d z}+\frac{4}{z} y=0. $$

Short answer

The general power series solution is \[ y(z) = a_0 + a_1z + \sum_{n=2}^{\infty} \frac{-4 a_{n-1}}{n(n-3)}z^n \]

Step-by-step solution

- Assume a power series solution

Assume that the solution can be written as a power series: \[ y(z) = \sum_{n=0}^{\infty} a_n z^n \]

- Compute the derivatives

Compute the first and second derivatives of the assumed power series:\[ y'(z) = \sum_{n=1}^{\infty} n a_n z^{n-1} \]\[ y''(z) = \sum_{n=2}^{\infty} n (n-1) a_n z^{n-2} \]

- Substitute into the differential equation

Substitute the power series and its derivatives back into the differential equation:\[ z \left(\sum_{n=2}^{\infty} n (n-1) a_n z^{n-2}\right) + (2z-3)\left(\sum_{n=1}^{\infty} n a_n z^{n-1}\right) + \frac{4}{z} \left(\sum_{n=0}^{\infty} a_n z^n\right) = 0 \]

- Simplify and combine like terms

Simplify each term and combine like powers of \(z\). You should get:\[ \sum_{n=2}^{\infty} n (n-1) a_n z^{n-1} + 2 \sum_{n=1}^{\infty} n a_n z^n - 3 \sum_{n=1}^{\infty} n a_n z^{n-1} + 4 \sum_{n=0}^{\infty} a_n z^{n-1} = 0 \]

- Adjust indices to combine series

Adjust the indices of the summations so that all terms are in the same power of \(z\) for comparison:\[ \sum_{n=1}^{\infty} (n(n-1) + 2n - 3n + 4a_{n-1}) z^{n-1} = 0 \]

- Form the recurrence relation

From the simplified series, form the recurrence relation for the coefficients by setting the coefficient of \[ a_n = \frac{-4a_{n-1}}{n(n-3)} \]

- Solve the recurrence relation

Solve the recurrence relation to find the general form of the coefficients.For \(a_0\):\[ a_0 = a_0 \text{ (free parameter)} \]For \(a_1\):\[ a_1 = a_1 \text{ (free parameter)}\]For higher coefficients:\[ a_n = \frac{-4a_{n-1}}{(n)(n-3)} \]

- Write the general power series solution

Substitute the values of the coefficients back into the original power series:\[ y(z) = a_0 + a_1z + \sum_{n=2}^{\infty} \frac{-4 a_{n-1}}{(n)(n-3)}z^n \]

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are essential for modeling phenomena where change occurs, such as in physics, engineering, and economics.
In this exercise, we deal with a second-order linear differential equation, meaning it involves the second derivative of an unknown function. The power series method is particularly useful because it helps find solutions even when traditional methods fall short. By expressing the solution as a power series, we assume that our function can be written in the form of an infinite sum of terms, like this:
\ y(z) = \sum_{n=0}^{\infty} a_n z^n \
Here, \( a_n \) are coefficients we aim to determine. This approach can handle more complex differential equations, including those with varying coefficients, like the one given. Power series solutions help simplify and manage the complexity by breaking down the problem into more manageable steps.

Recurrence Relations

Recurrence relations arise naturally when solving differential equations through the power series method. They provide a way to determine each coefficient in the series from the previous ones.
In the solution process, after substituting the power series and its derivatives back into the differential equation, we end up with a series that must equal zero. To satisfy this equality, the coefficients of each power of \( z \) must individually sum to zero. This requirement yields a recurrence relation:
\ a_n = \frac{-4a_{n-1}}{n(n-3)} \
This relation helps us compute each \( a_n \) step by step, starting from initial values like \( a_0 \) and \( a_1 \), which are free parameters. By iteratively applying the recurrence relation, we can build the complete solution for the differential equation.

Series Expansion

Series expansion is a fundamental technique in calculus and analysis, where a function is expressed as an infinite sum of terms. In the context of solving differential equations, these expansions—known as power series—allow us to write solutions that can handle complex variable interactions.
The general form of a power series expansion is:
\ y(z) = \sum_{n=0}^{\infty} a_n z^n \
where \( a_n \) are the coefficients determined through methods like recurrence relations. In our exercise, after solving for the \( a_n \) terms, we substitute them back into this general form to obtain the solution:
\ y(z) = a_0 + a_1z + \sum_{n=2}^{\infty} \frac{-4 a_{n-1}}{(n)(n-3)} z^n \
This series effectively represents the solution to the original differential equation. Series expansions are valuable because they can provide highly accurate approximations to functions and their behavior over an interval or around a point like \( z=0 \).