Question

Find the power series in \(x\) for the general solution. $$ \left(1-x^{2}\right) y^{\prime \prime}-5 x y^{\prime}-4 y=0 $$

Short answer

#Answer# The general solution of the given differential equation is $$ y(x) = \sum_{k=0}^{\infty} a_kx^k, $$ with the coefficients $a_k$ determined recursively through the relation $$ a_{k+2} = \frac{k(k-1)+5k+4}{(k+2)(k+1)}a_k. $$

Step-by-step solution

Assume a power series solution

Assume that we can represent the solution, y(x), as a power series with unknown coefficients, \(a_k\), as follows: $$ y(x) = \sum_{k=0}^{\infty} a_k x^k $$

Find the derivatives

In order to substitute the power series for y in the given equation, we need to find the first and second derivatives of y(x): $$ y'(x) = \sum_{k=1}^{\infty} ka_k x^{k-1} \\ y''(x) = \sum_{k=2}^{\infty} k(k-1)a_k x^{k-2} $$

Substitute the power series in the equation

Now, we substitute the series representation of y(x) and its derivatives into the given equation: $$ \left(1-x^2\right)\left(\sum_{k=2}^{\infty} k(k-1)a_k x^{k-2}\right) -5x\left(\sum_{k=1}^{\infty} ka_kx^{k-1}\right) - 4\left(\sum_{k=0}^{\infty} a_k x^k\right)=0 $$

Manipulate the series equation

Simplify the expression by distributing the coefficients in each series and reindexing the terms: $$ \left( \sum_{k=2}^{\infty} k(k-1)a_k x^{k-2} - \sum_{k=4}^{\infty} k(k-1)a_k x^{k} \right) -\sum_{k=1}^{\infty} 5ka_kx^{k} - \sum_{k=0}^{\infty} 4a_k x^k =0 $$ Now, shift indices in the first and second series to have the same power of x in each term: $$ \sum_{k=0}^{\infty} (k+2)(k+1)a_{k+2} x^{k} - \sum_{k=2}^{\infty} k(k-1)a_k x^{k} - \sum_{k=1}^{\infty} 5ka_kx^{k} - \sum_{k=0}^{\infty} 4a_k x^k =0 $$

Collect like terms

We now group the terms with like powers of x: $$ \sum_{k=0}^{\infty} [ (k+2)(k+1)a_{k+2} - k(k-1)a_k -5ka_k - 4a_k] x^{k} = 0 $$

Solve for the coefficients

Since the equation must hold for all powers of x, we can equate the coefficients of like powers of x to zero, and solve for the \(a_k\)'s: $$ (k+2)(k+1)a_{k+2} - k(k-1)a_k-5ka_k - 4a_k = 0 $$ Now, solving for \(a_{k+2}\), we get an expression for each of the coefficients as a function of the previous coefficients: $$ a_{k+2} = \frac{k(k-1)+5k+4}{(k+2)(k+1)}a_k $$

Determine the general solution

Using the recurrence relation for the coefficients, we can determine the general power series solution for the given differential equation: $$ y(x) = \sum_{k=0}^{\infty} a_kx^k $$ With the coefficients \(a_k\) determined recursively through the relation obtained in Step 6.

Key concepts

Power Series Solution

A power series is an infinite sum used to represent functions using terms based on increasing powers of a variable, in this case, the variable is \(x\). The expression for the power series is given by:

• \( y(x) = \sum_{k=0}^{\infty} a_k x^k \)

This representation assumes we can solve differential equations by expressing the unknown function as a series of powers of \(x\), with the coefficients \(a_k\) to be determined.
The process starts by writing the solution \(y(x)\) as a power series.

• This requires finding its derivatives, which are represented similarly as infinite sums.
• This makes it possible to substitute them back into the differential equation.

For a successful power series solution, these coefficients must satisfy the expanded differential equation. By using the power series expansion, the entire equation can be expressed in terms of corresponding powers of \(x\), allowing for an elegant solution method for complex differential equations.

Differential Equation Solution

Differential equations describe relationships involving derivatives, which indicate rates of change. Solving a differential equation means finding a function that satisfies this relation. Consider the differential equation:

• \( \left(1-x^{2}\right) y^{\prime \prime}-5 x y^{\prime}-4 y=0 \)

To solve this equation, we assumed a power series solution for \(y(x)\) and calculated its derivatives. By substituting the series representations of \(y(x)\), \(y'(x)\), and \(y''(x)\) into the equation, we can manipulate and rearrange the terms, collecting like terms together.
The solution involves equating coefficients of like powers of \(x\) to zero, ensuring that this sum holds true for all values of \(x\). This results in a recurrence relation that can be solved to find the series coefficients, \(a_k\), hence constructing the solution. The solution must balance out the equation for all terms to achieve a general solution.

Series Representations in Calculus

In calculus, series representations are powerful tools for solving equations, including differential equations. They provide a way to express complex functions as sums based on simpler components. Generally, a function \(y(x)\) can be represented as a series:

• \( y(x) = \sum_{k=0}^{\infty} a_k x^k \)

Using series in calculus, particularly with Taylor and power series, is beneficial because:

• They simplify the solving processes of intricate differential equations.
• Series allow functions to be analyzed term-by-term, offering clear insights and solutions.

A series representation approximates a function by summing up infinite terms where each is associated with a particular power of \(x\). The utility in differential calculus comes from simplifying complex functions into manageable components, easing both computation and conceptual understanding.