Question

In Exercises 33-46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution. $$ x^{2}\left(2+x^{2}\right) y^{\prime \prime}+x\left(3+x^{2}\right) y^{\prime}-y=0 $$

Short answer

Answer: The two fundamental Frobenius solutions are: Solution 1: $$ y_1(x) = a_0\sum_{n=0}^\infty \frac{(2n)!}{2^{2n}(n!)^2}x^{2n} $$ Solution 2: $$ y_2(x) = a_1\sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1}(n!)^2}x^{2n+1} $$ where \(a_0\) and \(a_1\) are constants.

Step-by-step solution

Express the equation in standard form

Dividing the given equation by \(x^2(2+x^2)\), we obtain: $$ y''+\frac{3+x^2}{2+x^2}y'-\frac{1}{x^2(2+x^2)}y=0 $$

Write down the Frobenius series solution

We assume a Frobenius series solution of the form: $$ y(x) = \sum_{n=0}^\infty a_n x^{n+r} $$ where \(a_n\) are the coefficients and r is a parameter.

Plug the Frobenius series into the equation and find the coefficients relation

Taking the derivatives of the series solution function, we get: $$ y'(x) = \sum_{n=0}^\infty (n + r) a_n x^{n+r-1} $$ and $$ y''(x) = \sum_{n=0}^\infty (n + r)(n + r - 1) a_n x^{n+r-2} $$ Now, substitute \(y(x)\), \(y'(x)\), and \(y''(x)\) into the standard form differential equation and rearrange the terms to obtain an equation for the coefficients: $$ \sum_{n=0}^\infty [(n + r)(n + r - 1) a_n x^{n+r-2}]+\frac{3+x^2}{2+x^2}\sum_{n=0}^\infty (n + r) a_n x^{n+r-1}-\frac{1}{x^2(2+x^2)}\sum_{n=0}^\infty a_n x^{n+r} = 0 $$

Solve for the coefficients

Because all the powers of x must cancel, we obtain two recurrence relations: 1. When n = 0: $$ r(r-1)a_0 = 0 $$ 2. When n > 0: $$ (n+r)(n+r-1)a_n + \frac{3}{2}(n+r)a_n - \frac{1}{2(n-1)}a_{n-2} = 0 $$ Using the first recurrence relation, we can solve for r: $$ r = 0 \text{ or } r = 1 $$ Now, using the second recurrence relation, we can find explicit formulas for the coefficients starting from \(n=2\): For \(r=0\): $$ a_{n}=\frac{a_{n-2}}{2n(2n-1)},\quad n\geq2, r=0 $$ For \(r=1\): $$ a_{n}=\frac{a_{n-2}}{2n(2n+1)},\quad n\geq2, r=1 $$ From here, we can build both solutions for the fundamental set of Frobenius solutions by plugging the expressions of the coefficients back to the general Frobenius series: Solution 1: $$ y_1(x) = a_0\sum_{n=0}^\infty \frac{(2n)!}{2^{2n}(n!)^2}x^{2n} $$ Solution 2: $$ y_2(x) = a_1\sum_{n=0}^\infty \frac{(2n)!}{2^{2n+1}(n!)^2}x^{2n+1} $$ Both solutions here form the fundamental set of Frobenius solutions.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They help us understand how quantities change and are used in various scientific fields, including physics and engineering. The key in differential equations is to find the function that satisfies the equation.
For the exercise provided, we have a second-order differential equation involving the function \(y\) and its derivatives \(y'\) and \(y''\). This equation is:
\[ x^{2}(2+x^{2})y'' + x(3+x^2)y' - y = 0 \]Understanding how to approach such types of equations involves knowing about series solutions, which are techniques to find functions that satisfy these equations.

Series Solutions

Series solutions refer to expressing the solution of a differential equation as an infinite sum of terms. A common type is the Frobenius series, which is particularly useful for solving linear differential equations near a singular point. This approach uses a series of the form:
\[ y(x) = \sum_{n=0}^\infty a_n x^{n+r} \]Here, \(a_n\) are coefficients that are determined through substitution into the differential equation, and \(r\) is a parameter that often represents the order of the singularity at a particular point.
Using this method in the given exercise, we assume that the solution can be expressed as a series, which allows us to handle the complexity of a differential equation by breaking it down into simpler, more manageable series terms.

Recurrence Relations

Recurrence relations are equations that recursively define sequences – each term is a function of preceding terms. In the context of series solutions for differential equations, they emerge naturally when substituting the series into the differential equation.
For our exercise, once the Frobenius series and its derivatives are substituted back into the differential equation, we need to equate coefficients of like powers of \(x\) to zero. This process generates a system of equations which are the recurrence relations for this problem:
- Initial condition: \(r(r-1)a_0 = 0\)- For \(n > 0\): \[ (n+r)(n+r-1)a_n + \frac{3}{2}(n+r)a_n - \frac{1}{2(n-1)}a_{n-2} = 0 \] These relations are essential for computing the coefficients \(a_n\) in the series solution.

Fundamental Set

A fundamental set of solutions for a differential equation refers to a set of linearly independent solutions that span the solution space of the equation. In simpler terms, any other solution to the equation can be written as a combination of the solutions in this set.
When a second-order differential equation is solved using the Frobenius method, you often find two linearly independent solutions, which together form this fundamental set. For our exercise, the solution set comprises two series:
- Solution 1: \(y_1(x)\)- Solution 2: \(y_2(x)\)These two solutions, when combined, can express any possible solution to the original differential equation.

Coefficients Formula

The coefficients formula provides explicit expressions for the coefficients \(a_n\) in the series solution of a differential equation. These are crucial as they define the terms of the series that approximate the solution.
For the exercise, after solving the recurrence relations, we find two coefficients formulas based on different values of \(r\):

• For \(r = 0\): \[ a_{n} = \frac{a_{n-2}}{2n(2n-1)}, \quad n \geq 2 \]
• For \(r = 1\): \[ a_{n} = \frac{a_{n-2}}{2n(2n+1)}, \quad n \geq 2 \]

These formulas are used iteratively starting from a known coefficient like \(a_0\) or \(a_1\), to calculate subsequent coefficients and build the series solutions \(y_1(x)\) and \(y_2(x)\), which form the fundamental set.