Chapter 7: Problem 7
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
Expert verified
#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
01
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:
$$
y(x) = \sum_{k=0}^{\infty} a_k x^k
$$
02
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):
$$
y'(x) = \sum_{k=1}^{\infty} ka_k x^{k-1} \\
y''(x) = \sum_{k=2}^{\infty} k(k-1)a_k x^{k-2}
$$
03
Substitute the power series in the equation
Now, we substitute the series representation of y(x) and its derivatives into the given equation:
$$
\left(1-x^2\right)\left(\sum_{k=2}^{\infty} k(k-1)a_k x^{k-2}\right) -5x\left(\sum_{k=1}^{\infty} ka_kx^{k-1}\right) - 4\left(\sum_{k=0}^{\infty} a_k x^k\right)=0
$$
04
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:
$$
\sum_{k=0}^{\infty} (k+2)(k+1)a_{k+2} x^{k} - \sum_{k=2}^{\infty} k(k-1)a_k x^{k} - \sum_{k=1}^{\infty} 5ka_kx^{k} - \sum_{k=0}^{\infty} 4a_k x^k =0
$$
05
Collect like terms
We now group the terms with like powers of x:
$$
\sum_{k=0}^{\infty} [ (k+2)(k+1)a_{k+2} - k(k-1)a_k -5ka_k - 4a_k] x^{k} = 0
$$
06
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:
$$
a_{k+2} = \frac{k(k-1)+5k+4}{(k+2)(k+1)}a_k
$$
07
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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:
The process starts by writing the solution \(y(x)\) as a power series.
- \( y(x) = \sum_{k=0}^{\infty} a_k x^k \)
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.
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:
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.
- \( \left(1-x^{2}\right) y^{\prime \prime}-5 x y^{\prime}-4 y=0 \)
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:
- \( y(x) = \sum_{k=0}^{\infty} a_k x^k \)
- They simplify the solving processes of intricate differential equations.
- Series allow functions to be analyzed term-by-term, offering clear insights and solutions.