Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}+8 x y^{\prime}-4 y=0 $$

Short answer

We solve the given differential equation, \(y'' + 8xy' - 4y=0\), using the power series method. First, we represent y, y', and y'' in power series form and substitute these expressions into the differential equation. After simplification and equating coefficients of like powers of x, we obtain the general recursion relation: \(a_{n+2} = \cfrac{-8(n+1)a_{n+1} + 4a_n}{(n+2)(n+1)}\) Computing even and odd coefficients using the recurrence relation and initial conditions, we find the power series solution: \(y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...\) This power series representation gives us an approximate solution to the given differential equation.

Step-by-step solution

Write the power series representation of y, y', and y''

First, let's find the power series representations of y, y', and y'': \(y(x) =\sum_{n=0}^{\infty} a_nx^n\) \(y'(x) =\sum_{n=1}^{\infty} na_nx^{n-1}\) \(y''(x) =\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}\)

Substitute the power series expressions into the differential equation

Now, we will substitute the power series of y, y', and y'' into the given differential equation, \(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + 8x\sum_{n=1}^{\infty} na_nx^{n-1} - 4\sum_{n=0}^{\infty} a_nx^n = 0\)

Simplify the equation

In this step, we will simplify the equation by multiplying the terms inside the sums and then combining the sums: \(\sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} + \sum_{n=1}^{\infty} 8na_nx^n - \sum_{n=0}^{\infty} 4a_nx^n = 0\) Now, we will change the index of summation to make all power series start from n=0. \(x^2\sum_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^n +\sum_{n=0}^{\infty} 8(n+1)a_{n+1}x^n - \sum_{n=0}^{\infty} 4a_nx^n = 0\)

Equate coefficients of like powers of x

Now, we can write the equation in general form by equating coefficients of like powers of x: \(x^2((n+2)(n+1)a_{n+2}) + 8(n+1)a_{n+1} - 4a_n = 0\) Solving for the general recursion relation: \((n+2)(n+1)a_{n+2} = -8(n+1)a_{n+1} + 4a_n\) Divide both sides by (n+2)(n+1): \(a_{n+2} = \cfrac{-8(n+1)a_{n+1} + 4a_n}{(n+2)(n+1)}\)

Compute the coefficients recursively

We can now compute the even and odd coefficients using our recurrence relation: For even terms, set n=2k: \(a_{2k+2} = \cfrac{-8(2k+1)a_{2k+1} + 4a_{2k}}{(2k+2)(2k+1)}\) For odd terms, set n=2k+1: \(a_{2k+3} = \cfrac{-8(2k+2)a_{2k+2} + 4a_{2k+1}}{(2k+3)(2k+2)}\) Using the initial conditions, we can compute the coefficients by starting with \(a_0\) and \(a_1\), and then using the recursion relation for subsequent coefficients.

Write the power series solution

Finally, we can write our power series solution as: \(y(x) = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...\) The power series representation gives us an approximate solution to the given differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In simple terms, they describe how a particular quantity changes over time or space. They are crucial in various fields of science and engineering for modeling real-world phenomena.
To solve a differential equation, we typically seek a function that satisfies the equation. However, finding exact solutions can be complicated, especially for non-linear differential equations. That's when alternative methods, like power series solutions, become handy.

• Linear vs Non-linear: Linear differential equations contain derivatives and terms that are proportional to the unknown function, while non-linear ones have more complex terms.
• Order: The order of a differential equation is the highest derivative it involves.

Recurrence Relations

Recurrence relations are equations that define sequences of numbers using previous terms. In the context of power series solutions, they are used to express the coefficients of the series in terms of one another, allowing for the recursive computation of each coefficient. This method simplifies finding the power series solution of a differential equation.
For instance, in our differential equation, we obtained: \[a_{n+2} = \frac{-8(n+1)a_{n+1} + 4a_n}{(n+2)(n+1)}\]This relation helps calculate all coefficients by using initial conditions and the recursion step.

• Base Case: You need a starting point, called initial conditions, to begin the recursive process.
• Recursive Step: The formula that relates unknown terms to known terms.

Power Series

A power series is essentially an infinite sum of terms in the form of \(a_n x^n\), where \(a_n\) are the coefficients and \(x\) is the variable. This concept is used to approximate more complex functions and is particularly useful when solving differential equations.

• Convergence: Power series converge within a certain radius, providing valid approximations only within this region.
• Representation: Instead of exact values, we often get approximations, which are useful in many applications like physics.

In the context of differential equations, power series solutions provide a method to build a function piece-by-piece by determining each term's coefficient through a recurrence relation.

Initial Conditions

Initial conditions are essential in solving differential equations as they provide the specific values needed to determine a unique solution for the problem at hand. They are the starting point for any recursion process used to determine coefficients in a power series solution.
For this differential equation, initial conditions typically involve specifying values like \(y(0)\) or \(y'(0)\). These conditions help us compute the coefficients \(a_0\), \(a_1\), and subsequent terms in the series using recurrence relations.

• Uniqueness: Initial conditions ensure that the solution to a differential equation is unique.
• Computation: They allow you to "kick-off" the process of computing a power series by starting with known values.