Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. $$ x^{2}\left(1+x^{2}\right) y^{\prime \prime}-x\left(7-2 x^{2}\right) y^{\prime}+12 y=0 $$

Short answer

To summarize, we used the Frobenius method to find two linearly independent solutions for the given linear, homogeneous, second-order differential equation. The method involved assuming a power series solution, finding its derivatives, and substituting them back into the differential equation. By equating the coefficients of each power of x, we found the recurrence relation for the coefficients of the series. Then, we used the recurrence relation to obtain an explicit formula for the coefficients and wrote our fundamental set of Frobenius solutions as: $$ y_{1}(x)=a_0 x^r \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k} $$ and $$ y_{2}(x)=a_1 x^{r+1} \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k} $$

Step-by-step solution

Write down the Frobenius series

Let's assume that the solution to the given differential equation can be expressed using the Frobenius method. The solution takes the form: $$ y(x) = x^r \sum_{n=0}^\infty a_n x^n $$

Differentiate the Frobenius series

To apply the Frobenius method, we first need to compute the first and second derivatives of the assumed solution \(y(x)\). We get: First derivative: $$ y'(x) = \sum_{n=0}^\infty (n+r)a_n x^{n+r-1} $$ Second derivative: $$ y''(x) = \sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2} $$

Substitute the derivatives into the given equation

Now, substitute the derivatives found in step 2, namely \(y'(x)\) and \(y''(x)\) back into the given original differential equation: $$ x^{2}\left(1+x^{2}\right) \left(\sum_{n=0}^\infty (n+r)(n+r-1)a_n x^{n+r-2}\right)-x\left(7-2 x^{2}\right) \left( \sum_{n=0}^\infty (n+r)a_n x^{n+r-1}\right)+12 \left( x^r \sum_{n=0}^\infty a_n x^n\right)=0 $$

Equate coefficients

Since the equation must hold for all values of x, equate the coefficients for each power of x in the sum: $$ \sum_{n=0}^\infty \left[(n+r)(n+r-1)a_n x^{n+r} - 7(n+r)a_n x^{n+r} + 2(n+r)a_n x^{n+r+2} + 12a_n x^{n+r}\right] = 0 $$ Now, for each power of x, set the coefficients equal to zero to find the recurrence relation for the coefficients \(a_n\).

Find the recurrence relation

After equating coefficients and simplifying, we obtain the following recurrence relation: $$ a_{n+2} = \frac{(n+r)(n+r-1) - 7(n+r) + 12}{(n+2)(n+r+2)}a_n $$ This recurrence relation will help us determine the coefficients \(a_n\) for the Frobenius series solution.

Find explicit formula for coefficients

We can now use the recurrence relation obtained in step 5 to find an explicit formula for the coefficients of the Frobenius solution. By repeatedly applying the recurrence relation and simplifying, we get: $$ a_n = \begin{cases}\frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)}a_0 & \text{for} \quad n=2k\\ \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)}a_1 & \text{for} \quad n=2k+1 \end{cases} $$ Now the two linearly independent Frobenius series solutions can be written as: $$ y_{1}(x)=a_0 x^r \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k} $$ and $$ y_{2}(x)=a_1 x^{r+1} \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k} $$ These are the fundamental set of Frobenius solutions with explicit formulas for the coefficients as required.

Key concepts

Differential Equations

A differential equation is a mathematical equation that involves functions and their derivatives. It relates a function to its rates of change and often arises from natural sciences describing real-world phenomena. The goal is often to find a function, or set of functions, that satisfies the equation for all points within a particular domain.
In the case presented, the differential equation is:

• \(x^{2}(1+x^{2}) y'' - x(7-2x^{2}) y' + 12 y = 0\)

This equation is second-order because the highest derivative is of second-degree, with variable coefficients making it more complex to solve directly.
Solutions often involve methods like separation of variables, integrating factor, or series solutions when the equation does not permit simple formulas.

Series Solution

A series solution is a technique applied to solve differential equations by assuming that the solution can be expressed as an infinite sum, which often takes the form of a power series or Frobenius series. This approach is particularly useful when the differential equation has variable coefficients that make traditional methods less effective or applicable.
The general form of the Frobenius series solution is:

• \(y(x) = x^r \sum_{n=0}^\infty a_n x^n\)

Here, \(x^r\) represents a potential characteristic exponent addressing singular points in the equation, and \(a_n\) are the coefficients to be determined by substitution into the differential equation.
Series solutions are powerful because they can transform a complex problem into a more manageable algebraic setup by determining the unknown coefficients through substitution and solving the resulting recurrence relation.

Recurrence Relation

A recurrence relation is an equation that expresses the coefficients of a series solution in terms of one or more of the preceding coefficients. In the context of solving differential equations with the Frobenius method, finding a recurrence relation allows you to compute every coefficient of the series based on the first few known terms.
From the problem's step-by-step solution, the recurrence relation is derived from the equated coefficients:

• \(a_{n+2} = \frac{(n+r)(n+r-1) - 7(n+r) + 12}{(n+2)(n+r+2)}a_n\)

This particular relation provides a systematic way to calculate subsequent values of \(a_n\), which is essential for building the Frobenius solution iteratively.
Recurrence relations simplify the solution construction by reducing it to repetitive computations which can easily be handled through a loop or recursion in programming.

Linear Independence

Linear independence is a crucial concept in the context of solutions to differential equations, especially when determining a fundamental set of solutions. In general, two functions are linearly independent if no nontrivial combination of them results in the zero function over a specific interval.
For the differential equation given, it results in two linearly independent Frobenius solutions:

• \(y_{1}(x)=a_0 x^r \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k}\)
• \(y_{2}(x)=a_1 x^{r+1} \sum_{k=0}^\infty \frac{(-1)^k (r+2k)(r+2k-1)}{(2k+1)(2k)(2)} x^{2k}\)

These solutions are linearly independent, which means that no scalar multiplication or addition of one can construct the other.
When you have two linearly independent solutions, you can form a general solution as a linear combination of both, providing a complete solution to the differential equation.