Problem 19 - Differential Equations and Linear Algebra

Question

(a) Determine a series solution to the initial-value problem $$4 y^{\prime \prime}+x y^{\prime}+4 y=0, \quad y(0)=1, \quad y^{\prime}(0)=0$$ (b) Find a polynomial that approximates the solution to Equation (11.2.19) with an error less than \(10^{-5}\) on the interval [-1,1] [Hint: The series obtained is a convergent alternating series.

Step-by-Step Solution

Answer

The series solution to the given initial-value problem is \(y(x) \approx 1 - \frac{1}{2}x^2 + \frac{1}{16}x^4\), which approximates the solution with an error less than \(10^{-5}\) on the interval [-1,1].

Step 1: Write down the given equation and its initial conditions

The given equation is \[4y''+xy'+4y=0\] with the initial conditions \(y(0)=1\) and \(y'(0)=0\).

Step 2: Introduce power series representations for the solution and its derivatives

Let \(y(x)\) be represented by a power series: \[y(x)=\sum_{n=0}^{\infty}a_nx^n\] Then we can find the derivatives of y(x), which are: \[y'(x)=\sum_{n=1}^{\infty}na_nx^{n-1}\] and \[y''(x)=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}\]

Step 3: Substitute the power series representations into the differential equation

Now, we plug the power series representations of \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given equation: \[4\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}+x\sum_{n=1}^{\infty}na_nx^{n-1}+4\sum_{n=0}^{\infty}a_nx^n=0\]

Step 4: Find a recurrence relation for coefficient a_n

First, we rewrite the summation terms \[\sum_{n=2}^{\infty}4n(n-1)a_nx^{n-2} = 4\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n\] and \[\sum_{n=1}^{\infty}xna_nx^{n-1} = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n+1}\] Substitute these summations back into the differential equation, and combine all of the terms under a single summation: \[\sum_{n=0}^{\infty}\left[4(n+2)(n+1)a_{n+2}+(n+1)a_{n+1}x+4a_n\right]x^n=0\] Equating each term in the summation to zero gives the recurrence relation \[4(n+2)(n+1)a_{n+2} +(n+1)a_{n+1}x +4a_n =0\]

Step 5: Solve the recurrence relation and use initial conditions

Rearrange the recurrence relation to find a formula for \(a_{n+2}\): \[a_{n+2}=-\frac{(n+1)a_{n+1}x+4a_n}{4(n+2)(n+1)}\] We use the initial conditions \(y(0)=1\) and \(y'(0)=0\) to set the first two coefficients, \(a_0 = 1\) and \(a_1 = 0\).

Step 6: Calculate the coefficients a_n iteratively

We can now calculate the coefficients by iterating the formula for \(a_{n+2}\) and use the fact that subsequent terms decrease alternately. By calculating few terms, we get \[a_2 = -\frac{4a_0}{4(2)(1)} = -\frac{1}{2}\] \[a_3 = 0\] \[a_4 = -\frac{4a_2}{4(4)(3)} = \frac{1}{16}\]

Step 7: Determine the polynomial solution

The error of the estimated solution is less than \(10^{-5}\) in the interval [-1,1]. We look for the smallest n where the term \(|a_nx^n|\) is smaller than \(10^{-5}\) for all \(x\in[-1, 1]\). We find that for n = 4, the term \(|a_4x^4|\) is smaller than \(10^{-5}\) in the interval [-1,1]. Hence, the polynomial solution approximating the solution with given error criterion is \[y(x) \approx a_0 + a_2x^2 + a_4x^4 = 1 - \frac{1}{2}x^2 + \frac{1}{16}x^4\]

Key Concepts

Initial-Value Problems

Initial-value problems are widely used in differential equations to find a specific solution that satisfies given conditions. These conditions, known as initial conditions, are usually applied at the beginning of the interval of interest. In the exercise, we have the second-order differential equation with initial conditions:

• Equation: \(4 y^{\prime \prime}+x y^{\prime}+4 y=0\)
• Initial Conditions: \(y(0)=1\) and \(y'(0)=0\)

These initial conditions help determine specific values for the coefficients of the power series solution, ensuring that the function not only satisfies the equation but also passes through the specified initial point on the graph. Without these conditions, we would have an infinite number of possible solutions that fit the equation. Initial conditions thus provide the needed constraints to ensure emantic and precise solutions.

Power Series Methods

Power series methods are powerful techniques used to solve differential equations, especially when traditional techniques prove to be cumbersome or inapplicable. These methods involve expressing the solution and its derivatives as infinite sums of power series, typically written in the form: \[y(x) = \sum_{n=0}^{\infty} a_n x^n\]The derivatives of this power series can then be computed term-wise, and substituted back into the given differential equation. For the exercise, we used the power series representation to substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the differential equation, which transforms it into a form that can be solved for the coefficients \(a_n\).This method is helpful when the solution to the differential equation isn't easily expressible in terms of familiar functions. By using power series, we gain approximation solutions that can provide insights into the behavior of the solution across an interval.

Polynomial Approximations

Polynomial approximations aim to simplify the power series solution into a more manageable form by considering only a finite number of terms. These approximations are especially useful over specified intervals where the power series converges quickly.In this exercise, we sought a polynomial approximation that would provide an error of less than \(10^{-5}\) within the interval [-1,1]. By examining the terms of the derived series, and finding that for \(n = 4\), the error threshold is met; we constructed the polynomial: \[y(x) \approx 1 - \frac{1}{2}x^2 + \frac{1}{16}x^4\]Polynomial approximations are beneficial for both practical computation and ease of graphing. They allow for a clear visual interpretation of the differential equation's solutions within the desired accuracy, facilitating better understanding and application within various contexts.

Alternating Series

Alternating series are sequences of terms that alternate in sign, typically converging more rapidly than non-alternating series. This makes them useful for approximating solutions to differential equations with greater efficiency.The alternating nature of the series in the exercise is evident from the coefficients calculated:

• \(a_2 = -\frac{1}{2}\)
• \(a_3 = 0\)
• \(a_4 = \frac{1}{16}\), and so on.

Due to their alternating signs, such series can be advantageous when assessing error bounds. Each subsequent term progressively decreases, offering insights into the convergence behavior and enabling approximations to meet stringent accuracy requirements like the one specified. Thus, understanding alternating series is crucial for effective error estimation and solution accuracy in differential equations.