Question

Find power series solutions in powers of \(x-1\) of each of the differential equations in Exercises. $$ \left(x^{4}-2 x^{3}+x^{2}\right) y^{\prime \prime}+2(x-1) y^{\prime}+x^{2} y=0 $$

Short answer

The power series solution for the given differential equation \(\left(x^{4}-2 x^{3}+x^{2}\right) y^{\prime \prime}+2(x-1) y^{\prime}+x^{2} y=0\) in powers of \(x-1\) is given by: \[y(x) =\sum_{n=0}^{\infty}a_n (x-1)^n = a_0 + a_1 (x-1) + a_2 (x-1)^2 + a_3 (x-1)^3 + \cdots\] where the coefficients \(a_n\) are computed iteratively using the recurrence relation: \[a_{n+2} = \frac{2n+1}{(n+2)(n+1)}a_{n+1} - \frac{x^2 - 1(x-1)}{(n+2)(n+1)}a_n\]

Step-by-step solution

Assume a Power Series Solution for \(y\)

We assume a power series solution for \(y\) in the form: \[y(x) = \sum_{n=0}^{\infty}a_n (x-1)^n\]

Derivatives of the Assumed Solution

Take the first and the second derivatives of \(y(x)\) and substitute them back into the original differential equation: \[y'(x) = \sum_{n=1}^{\infty} n a_n (x-1)^{n-1}\] \[y''(x) = \sum_{n=2}^{\infty} n (n-1)a_n (x-1)^{n-2}\] Now, substitute these derivatives back into the original equation: \[\left(x^{4}-2 x^{3}+x^{2}\right) \left(\sum_{n=2}^{\infty} n (n-1)a_n (x-1)^{n-2}\right) + 2(x-1)\left(\sum_{n=1}^{\infty} n a_n (x-1)^{n-1}\right) + x^{2}\left(\sum_{n=0}^{\infty}a_n (x-1)^n\right) = 0\]

Obtain Recurrence Relation for Coefficients

To identify a recurrence relation, we first need to find the coefficients of like terms. We can rewrite the above expressions with modified indices: \[\sum_{n=0}^{\infty} a_{n+2}n(n+1)(x-1)^n - 2 \sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1)(x-1)^n + \sum_{n=0}^{\infty} a_{n+2}(n+2)^2(x-1)^n + 2 \sum_{n=0}^{\infty} a_{n+1}(n+1)(x-1)^n + \sum_{n=0}^{\infty} a_n(x-1)^n (x^2 - 1(x-1))^n = 0\] Now, we equate the coefficients of like powers of \(x-1\) to zero to create the recurrence relation: \[a_{n+2}n(n+1) - 2a_{n+2}(n+2)(n+1) + a_{n+2}(n+2)^2 + 2a_{n+1}(n+1) + a_n(x^2 - 1(x-1))^n = 0\] Solving for \(a_{n+2}\), we find the recurrence relation: \[a_{n+2} = \frac{2n+1}{(n+2)(n+1)}a_{n+1} - \frac{x^2 - 1(x-1)}{(n+2)(n+1)}a_n\]

Construct Power Series Solutions

Using the recurrence relation obtained in step 3, we can compute coefficients and construct a power series solution for \(y(x)\). We can find the power series solution by iteratively solving the recurrence relation to obtain individual coefficients: \[y(x) =\sum_{n=0}^{\infty}a_n (x-1)^n = a_0 + a_1 (x-1) + a_2 (x-1)^2 + a_3 (x-1)^3 + \cdots\] By solving the recurrence relation for various values of \(n\), and inserting these coefficients back into the power series expansion, the desired power series solution in powers of \(x-1\) for the given differential equation can be obtained.

Key concepts

Differential Equations

Differential equations involve functions and their derivatives, representing complex physical systems or mathematical models. They contain equations that describe the rate of change of a quantity. In mathematical terms, a differential equation is an equation that establishes a relationship between a function and its derivatives. It takes the form:\[ F(x, y, y', y'', \ldots) = 0 \]Here, \( y \) is a dependent variable that depends on an independent variable \( x \), and \( y', y'' \), etc., are derivatives of \( y \) with respect to \( x \). Solving them means finding the function \( y(x) \) that satisfies the equation for particular conditions or domains. In this exercise, we have a second-order differential equation given:\[ \left(x^{4}-2 x^{3}+x^{2}\right) y^{\prime \prime}+2(x-1) y^{\prime}+x^{2} y=0 \]The goal is to find a power series solution around a specific point, empowering us to approximate solutions of such equations in terms of simple polynomial expressions.

Recurrence Relation

A recurrence relation is an equation that recursively defines a sequence of values. Each term in the sequence is defined as a function of its preceding terms. This method is particularly helpful in breaking down complex mathematical problems step-by-step.In the context of power series solutions for differential equations, a recurrence relation comes into play while determining coefficients of the power series.Once the derivatives of the assumed power series solution are substituted back into the differential equation, comparing the coefficients of the like terms leads to a recurrence relation. This relation helps to predict any term in the series based on previous terms. For the exercise, we derived the following recurrence relation:\[a_{n+2} = \frac{2n+1}{(n+2)(n+1)}a_{n+1} - \frac{x^2 - 1(x-1)}{(n+2)(n+1)}a_n\]This recurrence relation is crucial in systematically computing the coefficients \( a_n \) for successive powers of the series expansion.

Coefficients

Coefficients are numerical or constant multipliers attached to variables within algebraic expressions or equations. They determine the contribution of terms to an expression or function.In the power series solution of differential equations, the coefficients \( a_n \) are pivotal. They multiply each term of the power series, \[y(x) = \sum_{n=0}^{\infty}a_n (x-1)^n\]Finding the coefficients involves solving the derived recurrence relation for successive values of \( n \). For instance, starting with an initial value of \( a_0 \) and calculating subsequent coefficients \( a_1, a_2, \ldots \), these coefficients reconstruct the approximate solution of the function \( y(x) \). By calculating accurate coefficients, one can sketch a precise representation of the function within the range of interest.

Power Series Expansion

Power series expansion allows us to represent a function as an infinite sum of terms based on powers of a variable. This representation is especially useful for approximating functions in calculus and mathematical analysis.A power series around \( x-1 \) takes the form:\[y(x) = \sum_{n=0}^{\infty}a_n (x-1)^n\]In the given differential equation, we use power series to express the solution as a linear combination of polynomial terms. By expressing \( y, y' \), and \( y'' \) in the power series form and substituting back into the differential equation, we obtain individual equations for each power of \( (x-1) \).This approach allows for both the construction of approximate solutions and deeper insights into the behavior of differential equations. By solving for the coefficients sequentially using a recurrence relation, one can construct precise and useful representations of solutions, even in complex domains.