Question

Find \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in the power series \(y=\sum_{n=0}^{\infty} a_{n}\left(x-x_{0}\right)^{n}\) for the solution of the initial value problem. Take \(x_{0}\) to be the point where the initial conditions are imposed. $$ \left(5-6 x+3 x^{2}\right) y^{\prime \prime}+(x-1) y^{\prime}+12 y=0, \quad y(1)=-1, \quad y^{\prime}(1)=1 $$

Short answer

Question: Find the first 8 coefficients of the power series solution of the second-order linear differential equation \((5-6x+3x^2)y'' +(x-1)y' +12y=0\) with initial conditions \(y(1)=-1\) and \(y'(1)=1\). Answer: The first 8 coefficients of the power series solution are \(a_0 = -1\), \(a_1 = 1\), \(a_2 = 3\), \(a_3 = -\frac{1}{2}\), \(a_4 = -\frac{3}{2}\), \(a_5 = \frac{1}{8}\), \(a_6 = \frac{1}{10}\), and \(a_7 = -\frac{1}{84}\).

Step-by-step solution

Represent y, y', and y'' as power series

First, let's express the given y, y', and y'' as a power series. $$ y = \sum_{n=0}^{\infty} a_{n}(x-1)^n $$ Taking the derivative of y with respect to x: $$ y' = \sum_{n=1}^{\infty} na_{n}(x-1)^{n-1} $$ Taking second derivative of y with respect to x: $$ y'' = \sum_{n=2}^{\infty} n(n-1)a_{n}(x-1)^{n-2} $$

Substitute the power series expressions into the differential equation

Now, we need to substitute these power series representations of y, y', and y'' into the given differential equation: $$ (5-6x+3x^2)\left(\sum_{n=2}^{\infty} n(n-1)a_{n}(x-1)^{n-2}\right) +(x-1)\left(\sum_{n=1}^{\infty} n a_{n}(x-1)^{n-1}\right) + 12\left(\sum_{n=0}^{\infty} a_{n}(x-1)^n\right) = 0 $$

Find the recurrence relation

In this step, we'll match the coefficients of each power of (x-1) to find the recurrence relation. First, expand the differential equation expression: $$ \sum_{n=2}^{\infty} (5-6x+3x^2)n(n-1)a_n(x-1)^{n-2}+ \sum_{n=1}^{\infty} (x-1)n a_n(x-1)^{n-1}+ \sum_{n=0}^{\infty} 12a_n(x-1)^{n}=0 $$ Now, we'll multiply each term by \((x - 1)^2\) to simplify the expression and get the recurrence relation. $$ \sum_{n=2}^{\infty} (5-6x+3x^2)n(n-1)a_n(x-1)^n+ \sum_{n=1}^{\infty} (x-1)^3n a_n(x-1)^n+ \sum_{n=0}^{\infty} 12(x-1)^2a_n(x-1)^n=0 $$ By looking at the coefficients of each power of \((x-1)^n\), we can find the recurrence relation: $$ (5-6x_0+3x_0^2)n(n-1)a_n + (x_0-1)^3na_{n-1} + 12(x_0-1)^2a_{n-2} = 0 $$ Since \(x_0 =1\), the recurrence relation simplifies to: $$ 2n(n-1)a_n + 12a_{n-2} = 0 $$

Use the initial conditions to find the coefficients

Now, we will use the initial conditions to find the coefficients of the power series. $$ y(1)=-1 \Rightarrow \sum_{n=0}^{\infty} a_{n}(1-1)^n = -1 \Rightarrow a_0 = -1 $$ $$ y'(1)=1 \Rightarrow \sum_{n=1}^{\infty} na_{n}(1-1)^{n-1} = 1 \Rightarrow a_1 = 1 $$ Now, let's find the coefficients from \(a_2\) to \(a_7\) using the recurrence relation we derived in Step 3. We will use the formula \(a_n = -\frac{6}{n(n-1)}a_{n-2}\) to find the required coefficients. $$ a_2 = -\frac{6}{2(2-1)}a_0 = 3 $$ $$ a_3 = -\frac{6}{3(3-1)}a_1 = -\frac{1}{2} $$ $$ a_4 = -\frac{6}{4(4-1)}a_2 = -\frac{3}{2} $$ $$ a_5 = -\frac{6}{5(5-1)}a_3 = \frac{1}{8} $$ $$ a_6 = -\frac{6}{6(6-1)}a_4 = \frac{1}{10} $$ $$ a_7 = -\frac{6}{7(7-1)}a_5 = -\frac{1}{84} $$ Therefore, the coefficients are: \(a_0 = -1\), \(a_1 = 1\), \(a_2 = 3\), \(a_3 = -\frac{1}{2}\), \(a_4 = -\frac{3}{2}\), \(a_5 = \frac{1}{8}\), \(a_6 = \frac{1}{10}\), and \(a_7 = -\frac{1}{84}\).

Key concepts

Understanding Initial Value Problems

When tackling differential equations, an 'initial value problem' is an equation coupled with specified values, known as initial conditions, that the solution must satisfy at a given point. These problems are fundamental because they often represent physical phenomena where conditions are known at a specific instant, and our goal is to predict future behavior.

For the example at hand, we have an initial value problem where both the function and its first derivative are specified when the independent variable is equal to one: the function value (\(y(1) = -1\)) and the first derivative value (\(y'(1) = 1\)). This data greatly aids in determining unique solutions, ensuring that the resulting power series is tailored precisely to those initial conditions, leading to an explicit determination of the coefficients in the power series representation.

The Role of Recurrence Relations

Recurrence relations are vital tools in solving differential equations with power series methods. They provide a sequential approach to finding coefficients in the power series expansion by expressing each term in relation to previous ones.

By equating coefficients of corresponding powers after substituting the series into the differential equation, we unveil a pattern or a formula that relates the coefficients. For example, in our exercise, we derived a simple linear recurrence relation \(2n(n-1)a_n + 12a_{n-2} = 0\), which allows us to recursively calculate the coefficients \(a_n\) by starting with known initial values. This sequence becomes a roadmap, directing us from our initial conditions to the coefficients we need to solve the problem, turning a potentially complex process into manageable steps.

Power Series Representation of Solutions

In the realm of differential equations, power series representations are akin to spelling out solutions in an infinite sum of terms. They are particularly effective when dealing with non-polynomial equations, where traditional algebraic methods might fail.

In the exercise, our power series solution takes the form \(y = \(sum_{n=0}^{\infty} ) a_{n}\)(x - x_{0})^{n}\), where \(a_n\) are the coefficients we seek. This method hinges on the fact that the differential equation can be interpreted in terms of these series, providing a direct method to calculate an otherwise elusive solution. By finding a specified number of coefficients, we can sketch out the shape of the solution curve near the initial point, which in many practical scenarios, is sufficient to understand the behavior of the system being modeled.