Question

Find solutions, as power series in \(z\), of the equation $$ 4 z y^{\prime \prime}+2(1-z) y^{\prime}-y=0. $$ Identify one of the solutions and verify it by direct substitution.

Short answer

Assume a power series solution, find the recurrence relation, and determine terms.

Step-by-step solution

- Assume a Power Series Solution

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

- Compute the Derivatives

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

- Substitute into Differential Equation

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

- Combine and Simplify Series

Combine the series into a single sum and align terms by powers of z: \[ \sum_{n=0}^{\infty} [(4n(n-1) a_n z^n + 2n a_n z^{n-1} - 2n a_n z^n - a_n z^n] = 0 \]

- Collect Like Terms and Simplify

Align terms in powers of z from \( z^n \): \[4n(n-1) a_n + 2(n+1) a_{n-1} - a_n = 0 \]

- Find Recurrence Relation

Solve for the recurrence relation: \[a_n = \frac{2(n-1)a_{n-1}}{1 +4 (n-1)} \]

- Determine the Coefficients

Choose a_0 such that a_n is determined by solving the recurrence relation.

- Identify Particular Solution

Identify a particular solution by setting initial terms (for example, set \(a_0 = 1 \), \(a_1= 0 \) ). It will give the power series form for coefficients.

- Verify Solution by Direct Substitution

Verify by substituting: substitute \( y(z) = a_0 F(z) \) into the original equation and ensure it satisfies it.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They are essential for modeling various physical systems, such as heat transfer, motion of planets, and population dynamics.
In our exercise, we deal with a second-order linear differential equation of the form:
\ \[ 4 z y'' + 2(1 - z) y' - y = 0 \ \]
Here, y is a function of z, and y' and y'' are its first and second derivatives, respectively.

To solve this equation using a power series approach, we assume that the solution y(z) can be expressed as a power series:
\ \[ y(z) = \sum_{n=0}^{\infty} a_n z^n \ \]
Here, a_n represents the series coefficients that we need to determine. The key steps involve substituting this series into the differential equation, aligning terms of the same power of z, and solving for the coefficients.

Leverage this power series approach when you need an analytical solution that can be flexible in handling a wide range of initial and boundary conditions.

Recurrence Relation

A recurrence relation is an equation that expresses each term of a sequence as a function of the preceding terms. In our problem, we derive the recurrence relation by substituting the power series and its derivatives back into the differential equation:
\ \[ 4z \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} + 2(1 - z) \sum_{n=1}^{\infty} n a_n z^{n-1} - \sum_{n=0}^{\infty} a_n z^n = 0 \ \]
Combining and simplifying these series, we separate terms by powers of z, leading us to:
\sum_{n=0}^{\infty} \left[ 4n(n-1) a_n z^n + (2n a_n z^{n-1} - 2n a_n z^n) - a_n z^n \right] = 0.
Collecting terms of like powers in z and further simplifying, we get the recurrence relation:
\ \[ a_n = \frac{2(n-1) a_{n-1}}{1 + 4(n - 1)} \ \]
This relation helps us determine each coefficient a_n using the previous coefficient a_{n-1}. Setting the initial conditions and solving this relation step-by-step helps us identify the new terms in the power series.

Verification of Solutions

After obtaining the power series solution, we must verify it by direct substitution back into the original differential equation to ensure it is correct. Verification shows that our derived series truly satisfies the differential equation.

We start by taking our identified power series solution:
\ \[ y(z) = \sum_{n=0}^{\infty} a_n z^n \ \]
and its derivatives:
\ \[ y'(z) = \sum_{n=1}^{\infty} n a_n z^{n-1}, \ \]
\ \[ y''(z) = \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} \ \]
We substitute these into the original differential equation:
\ \[ 4 z y'' + 2(1 - z) y' - y = 0 \ \]
Simplifying the equation while ensuring it holds for all powers of z confirms that the provided power series is indeed a valid solution.

Verification not only proves our solution’s correctness but also strengthens our understanding of the problem-solving process. Different problems may require different verification steps, but the essence remains: check consistency with the original equation.