Problem 111 - Calculus 3

Question

Find a power series solution for the following differential equations. $$y^{\prime \prime}-x y^{\prime}-y=0$$

Step-by-Step Solution

Answer

The power series solution is \( y(x) = \sum_{n=0}^{ fty} \frac{x^{2n}}{(2n)!} \).

Step 1: Assume a Power Series Solution

Assume a power series solution for the function \( y(x) = \sum_{n=0}^{fty} a_n x^n \). The goal is to find expressions for the coefficients \( a_n \) such that this power series satisfies the given differential equation.

Step 2: Compute Derivatives

Compute the first and second derivatives of the assumed power series:- The first derivative is \( y'(x) = \sum_{n=1}^{fty} n a_n x^{n-1} \).- The second derivative is \( y''(x) = \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} \).

Step 3: Substitute into the Differential Equation

Substitute the power series expressions for \( y''(x) \), \( y'(x) \), and \( y(x) \) into the differential equation:\[ \sum_{n=2}^{fty} n(n-1) a_n x^{n-2} - x \sum_{n=1}^{fty} n a_n x^{n-1} - \sum_{n=0}^{fty} a_n x^n = 0. \]

Step 4: Match Powers of x

Align the series by matching powers of \( x \), which requires changing indices:- For \( y''(x) \), let \( m = n - 2 \), so \( \sum_{m=0}^{fty} (m+2)(m+1) a_{m+2} x^m \).- For \( x y'(x) \), let \( m = n - 1 \), so \( \sum_{m=0}^{fty} m a_m x^m \).- For \( y(x) \), it is \( \sum_{m=0}^{fty} a_m x^m \).

Step 5: Combine and Simplify the Series

Combine the re-indexed equations:\[ \sum_{m=0}^{fty} \left((m+2)(m+1) a_{m+2} - m a_m - a_m \right) x^m = 0. \]This simplifies to:\[ \sum_{m=0}^{fty} \left((m+2)(m+1) a_{m+2} - (m+1) a_m \right) x^m = 0. \]

Step 6: Solve Recurrence Relation

Since the sum equals zero for all \( x \), each coefficient must individually equal zero:\[ (m+2)(m+1) a_{m+2} = (m+1) a_m. \]This gives the recurrence relation:\[ a_{m+2} = \frac{a_m}{m+2}. \]

Step 7: Determine Initial Coefficients

Select initial coefficients:- \( a_0 \) is arbitrary (typically set to 1 for simplicity).- \( a_1 \) is also arbitrary (can be set to 0 if not specified in the problem).Using the recurrence relation:- \( a_2 = \frac{a_0}{2} \)- \( a_3 = \frac{a_1}{3} = 0 \) if \( a_1 = 0 \) as chosen.

Step 8: Construct the Power Series Solution

Using the initial conditions:\( y(x) = a_0 + a_1 x + \frac{a_0}{2} x^2 + \frac{a_1}{3} x^3 + \cdots \)With choices \( a_0 = 1 \) and \( a_1 = 0 \), this simplifies to:\[ y(x) = 1 + \frac{x^2}{2} + \frac{x^4}{4!} + \cdots = \sum_{n=0}^{fty} \frac{x^{2n}}{(2n)!}. \]

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates a function with its derivatives. In simple terms, it involves an unknown function and its derivatives. Differential equations are used to describe a wide array of phenomena such as growth rates, motion, and changes in physical systems.
In this context, the equation given is \[ y'' - x y' - y = 0, \] which is a second-order linear differential equation. It means that the highest derivative is the second derivative. The task is to find a power series solution for this differential equation. The power series solution approach turns this higher-order problem into solving for a sequence of coefficients.

Recurrence Relation

A recurrence relation is a mathematical relation that expresses the next term in a sequence using the preceding terms.
When solving differential equations through power series, the goal is to find these relations between the coefficients of the series.
In our step-by-step solution, we derived the following recurrence relation:\[ a_{m+2} = \frac{a_m}{m+2}. \] This equation tells us how to calculate each coefficient in terms of an earlier coefficient. By solving this relation, we can determine each successive coefficient based on our initial conditions.

Initial Coefficients

Determining initial coefficients is a crucial step in solving differential equations using power series. The initial coefficients provide the starting point for constructing the power series solution.
In this exercise, initial coefficients like \( a_0 \) and \( a_1 \) are selected arbitrarily to simplify calculations. Typically, \( a_0 \) is set to 1 and \( a_1 \) to 0, unless there are specific initial conditions provided by the problem.
These coefficients help build the entire solution by providing initial values for the recurrence relation:

• \( a_0 = 1 \)
• \( a_1 = 0 \)
• \( a_2 = \frac{a_0}{2} = \frac{1}{2} \)

Thanks to this process, the power series representing the solution can be constructed as:\[ y(x) = 1 + \frac{x^2}{2} + \frac{x^4}{4!} + \cdots. \]