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. $$ \left(2 x^{2}+4 x+5\right) y^{\prime \prime}-20(x+1) y^{\prime}+60 y=0, \quad y(-1)=3, \quad y^{\prime}(-1)=-3 $$

Short answer

Using the power series method, we found that the first eight coefficients of the power series for the given differential equation and initial conditions are: $$ a_0 = 3, \\ a_1 = 4, \\ a_2 = -1, \\ a_3 = 0, \\ a_4 = 0, \\ a_5 = 0, \\ a_6 = 0, \\ a_7 = 0. $$

Step-by-step solution

Substitute the power series into the given differential equation.

First, we'll need to represent \(y^{\prime}\) and \(y^{\prime\prime}\) as power series. Since \(y=\sum_{n=0}^{\infty} a_{n}(-1)^{n}(x+1)^{n}\), we have: $$ y^{\prime} = \sum_{n=1}^{\infty} na_{n}(-1)^{n}(x+1)^{n-1} \\ y^{\prime\prime} = \sum_{n=2}^{\infty} n(n-1)a_{n}(-1)^{n}(x+1)^{n-2}. $$ Now, substitute \(y^{\prime\prime}\), \(y^{\prime}\), and \(y\) into the given differential equation: $$ (2x^2 + 4x + 5) \left( \sum_{n=2}^{\infty} n (n-1)a_{n}(-1)^{n}(x+1)^{n-2} \right) \\ - 20(x+1) \left(\sum_{n=1}^{\infty} na_{n}(-1)^{n}(x+1)^{n-1} \right)\\ + 60 \left(\sum_{n=0}^{\infty} a_{n}(-1)^{n}(x+1)^{n} \right) = 0. $$

Equate coefficients of matching powers of \((x-x_{0})\)

To equate coefficients of matching powers of \((x-x_{0})\), we will need to multiply the power series terms by the coefficients of \(y^{\prime\prime}\), \(y^{\prime}\), and \(y\) in the given differential equation and combine the corresponding terms. $$ \sum_{n=2}^{\infty} n(n-1)a_{n}(2x^2 + 4x + 5)(x+1)^{n-2}\\ - 20 \sum_{n=1}^{\infty} na_{n}(x+1)^{n}\\ + 60 \sum_{n=0}^{\infty} a_{n}(x+1)^{n} = 0. $$ The sum of the coefficients for a given power of \((x+1)\) on both sides of the equation must be equal. Thus, we can write: $$ n(n-1)a_{n}(2x^2 + 4x + 5) = 20na_{n} - 60a_{n} \implies \\ a_{n} = \frac{20n - 60}{n(n-1)(2a_{n-2} + 4a_{n-1} + 5)}. $$

Solve for \(a_n\)

We can use the above equation to solve for the coefficients \(a_n\) iteratively: For \(n = 1\): $$ a_1 = \frac{20(1) - 60}{1(0)(2a_{-1} + 4a_{0} + 5)} = \frac{-40}{0} = \text{Undefined}. $$ However, since we derived the equation for \(a_n\) for \(n \ge 2\), we need to apply the initial conditions to find \(a_0\) and \(a_1\).

Apply the initial conditions

Let's use the initial conditions provided: $$ y(-1) = 3 = \sum_{n=0}^{\infty} a_{n}(-1)^{n}(x+1)^{n} \implies a_0 - a_1 + a_2 - \cdots = 3,\\ y'(-1) = -3 = \sum_{n=1}^{\infty} n a_{n}(-1)^{n}(x+1)^{n-1}\implies -a_1 + 2a_2 - 3a_3 + \cdots = -3. $$ From these equations, we know: $$ a_0 = 3, \\ a_1 = a_0 + 1 = 4. $$ Now, we can find the other coefficients by using the equation we derived in Step 2: For \(n = 2\): $$ a_{2} = \frac{20(2) - 60}{2(1)(2a_{0} + 4a_{1} + 5)} = \frac{-20}{10(13)} = -1. $$ For \(n = 3\): $$ a_{3} = \frac{20(3) - 60}{3(2)(2a_{1} + 4a_{2} + 5)} = \frac{0}{6(-8)} = 0. $$ Since \(a_3=0\), we can determine that each subsequent \(a_n\) will also be zero: For \(n > 3\): $$ a_{n} = \frac{20(n) - 60}{n(n-1)(2a_{n-2} + 4a_{n-1} + 5)} = \frac{20(n) - 60}{n(n-1)(0)} = 0. $$ Thus, the first eight coefficients of the power series are: $$ a_0 = 3, \\ a_1 = 4, \\ a_2 = -1, \\ a_3 = 0, \\ a_4 = 0, \\ a_5 = 0, \\ a_6 = 0, \\ a_7 = 0. $$

Key concepts

Power Series

A power series is a type of series in mathematics that represents a function as the sum of an infinite number of terms. Each term is multiplied by a constant and raised to the power of a variable. Specifically, the power series is an expansion of the form:

• \[ y = \sum_{n=0}^{\infty} a_{n}(x-x_{0})^{n} \]

The coefficients \(a_n\) determine the specific properties and behavior of the series. Power series are particularly useful for approximating functions and can often be used to solve differential equations. When applied to differential equations, the coefficients of the series are determined so that the series satisfies the equation everywhere within its radius of convergence.

Differential Equation Solution

Solving a differential equation involves finding a function or a set of functions that satisfies the equation for all points in a specified domain. Specifically for this problem, we are focusing on a second-order linear differential equation:

• \( (2x^2 + 4x + 5) y^{\prime\prime} - 20(x+1) y^{\prime} + 60 y = 0 \)

The solution process involves substituting the power series expressions for \(y\), \(y^{\prime}\), and \(y^{\prime\prime}\) into the differential equation and then equating coefficients for corresponding powers of \((x-x_{0})\). This approach converts the differential equation into a set of algebraic equations for the power series coefficients, \(a_n\). This method can be particularly effective when the differential equation has variable coefficients, and the solution requires calculating terms iteratively.

Taylor Series Coefficients

In the context of power series, Taylor series coefficients \(a_n\) are determined so that the series matches a function and its derivatives at a specific point \(x_0\). The Taylor series provides an approximation of functions using derivatives at a single point:

• \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \)

In our exercise, we are deriving the coefficients by substituting into the differential equation and applying initial or boundary conditions. The coefficients are calculated to ensure that the series solves the differential equation and meets the given conditions at \(x_0\) (e.g., the initial conditions provided). The coefficients must satisfy the equations derived from matching the series to the differential equation. This involves step by step calculation using the ratios derived from the series terms, as shown in the example provided.

Boundary Conditions

Boundary conditions are essential specifications required to ensure the uniqueness of the solution of a differential equation. They dictate the behavior of the solution at specific points, typically at the beginning or end of an interval. Initial value problems specifically use conditions at the start point to determine a unique solution.

• Example conditions such as \( y(-1)=3 \) and \( y^{\prime}(-1)=-3 \) provide necessary values for constants within the solution.

By plugging these conditions into the series representation of the solution, we find specific values for unknown coefficients such as \(a_0\) and \(a_1\). These boundary values lead to a complete solution that is specific and tailored to meet the given conditions. Their role ensures that the power series aligns with the expected behavior of the solution at these critical points.