Question

Find power series solutions in powers of \(x\) of each of the differential equations in Exercises. $$ y^{\prime \prime}-\left(x^{3}+2\right) y^{\prime}-6 x^{2} y=0 $$

Short answer

The power series solution for the given differential equation \(y^{\prime \prime}-\left(x^{3}+2\right) y^{\prime}-6 x^{2} y=0\) is in the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$ With the coefficients \(a_{n+2}\) determined recursively as: $$ a_{n+2} = \frac{(n+1)a_{n+1}+ 6 a_n }{(n+2)(n+1)} $$ The series will generally be expressed in terms of initial values, \(a_0\) and \(a_1\).

Step-by-step solution

Assume a General Power Series Form for the Solution

Assuming a power series solution for y(x) in the form of: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$ where \(a_n\) are the coefficients to be determined in the series.

Calculate the First and Second Derivatives

Now, find the first and second derivatives: $$ y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} $$ $$ y^{\prime\prime}(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} $$

Substitute Derivatives into the ODE

Substitute the expressions for y, y', and y'' into the given differential equation: $$ \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - \left(x^{3}+2\right)\sum_{n=1}^{\infty} n a_n x^{n-1} - 6x^2\sum_{n=0}^{\infty} a_n x^n = 0 $$

Expand and Collect the Terms

Now, expand the terms, and collect similar powers of x: \[ (x^2 + 2) \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2} - x^3\sum_{n=1}^{\infty} n a_n x^{n-1} - 6x^2\sum_{n=0}^{\infty} a_n x^n = 0 \] \[ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} (n+1) a_{n+1} x^{n+3} - 6\sum_{n=0}^{\infty} a_n x^{n+2} = 0 \] Redefine the summations, so they all start from n = 0 \[ \left[ \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n \right] - \left[x^3\sum_{n=0}^{\infty} (n+1) a_{n+1} x^{n}\right] - \left[6x^2\sum_{n=0}^{\infty} a_n x^{n}\right] = 0 \]

Equate the Coefficients for each Power of x

Because the power series on both sides of the equation must match for all x, we can equate the coefficients for each power of x: $$ (n+2)(n+1)a_{n+2} - (n+1)a_{n+1} - 6 a_n = 0 $$ This allows us to express each new coefficient in terms of earlier coefficients: $$ a_{n+2} = \frac{(n+1)a_{n+1}+ 6 a_n }{(n+2)(n+1)} $$

Solve for the Power Series Coefficients

Starting with n = 0 and solving for each coefficient recursively with the given formula, we can find the power series solution for the given ODE. The series will generally be expressed in terms of initial values, a_0 and a_1. The final solution is in the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^n $$ However, it is important to note that, practically, it's not possible to compute this sum for all possible terms, and we can only compute a finite number of terms to get an approximation of the solution.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving functions and their derivatives. They play a vital role in many areas of science and engineering, including mechanics, economics, and biology, describing how a certain quantity changes over time or space. The ODE in our exercise,
\(y'' - (x^3 + 2)y' - 6x^2y = 0\),
is a second-order linear homogeneous equation with variable coefficients. It's essential to understand that solving an ODE means finding a function (or set of functions) that satisfies the equation, and power series methods provide one such technique for finding these functions, especially when the ODE cannot be solved by elementary methods.

Power Series

A power series is an infinite sum of the form
\(y(x) = \sum_{n=0}^{\infty} a_n x^n\),
where \(a_n\) represents the coefficient of the nth term, and \(x^n\) represents the nth power of the variable x. In the context of solving differential equations, assuming that the solution can be expressed as a power series allows us to handle ODEs with non-constant coefficients by matching the terms of identical powers in x, leading to a series of equations that can be solved for the coefficients \(a_n\).

Derivatives in Power Series

Calculating derivatives in power series is straightforward. Each term of the series is differentiated according to the power rule, leading to a new power series. For example, the first derivative of a series takes the form:
\(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\),
and the second derivative is:
\(y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\).
Crucially, differentiating term by term is justified within the radius of convergence of the series. These derived series are used for substituting back into the original ODE, as we did in the example problem, allowing us to find a relationship between the series' coefficients.

Coefficient Recursion

Coefficient recursion is a technique for generating the next term in a series based on previous terms. In our differential equation example, we ended up with a recursive relationship:
\(a_{n+2} = \frac{(n+1)a_{n+1}+ 6 a_n }{(n+2)(n+1)}\).
This recursive formula allows us to calculate each coefficient systematically, starting from known initial values. The recursion dramatically simplifies the computation process, making it possible to determine as many coefficients as needed for a sufficient approximation of the solution.

Initial Value Problems

Initial value problems involve finding a specific solution to a differential equation that satisfies given initial conditions. For a second-order ODE like the one in our exercise, two initial values are typically required, often corresponding to the value of the function and its first derivative at some point, typically \(x = 0\). In the context of power series solutions, these initial values will correspond to \(a_0\) and \(a_1\), the first two coefficients of the series. By determining these values, we can then use the recursive formula to build out the series that is the solution to the ODE.