Question

In Exercises 33-46 find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients in each solution. $$ x\left(1+x^{2}\right) y^{\prime \prime}+\left(4+7 x^{2}\right) y^{\prime}+8 x y=0 $$

Short answer

Question: Find the fundamental set of Frobenius solutions for the given differential equation: $$y'' + \frac{4 + 7x^2}{x(1+x^2)}y' + \frac{8x}{x(1+x^2)}y = 0 $$. Answer: Based on the procedure we followed, the fundamental set of Frobenius solutions can be given as follows: $$ y(x)=c_1\left(a_0+\sum_{n=1}^{\infty} a_nx^{n+r}\right)+c_2\underbrace{\left(a_1+\sum_{n=1}^{\infty}\left[\frac{(n+r)(n+r-1)}{(4+7(n+r-1)^2)(n+r-1)(1+(n+r-1)^2)}\right] a_nx^{n+r}\right)}_{=b_1+a_{1,1}x^{r_1}+a_{2,1}x^{r_2}+\cdots}, $$ where \(c_1\) and \(c_2\) are constants, and the coefficients \(a_n\) are determined from the explicit formulas obtained in step 5.

Step-by-step solution

Rewriting the given equation in standard form

Rewrite the given equation as follows: $$ y^{\prime \prime}+\left(\frac{4+7 x^{2}}{x\left(1+x^{2}\right)}\right) y^{\prime}+\left(\frac{8 x}{x\left(1+x^{2}\right)}\right) y=0. $$ Notice that the equation is now in standard form for the Frobenius method.

Assuming a Frobenius solution in the form of a power series

We will assume a Frobenius solution in the form of a power series: $$ y(x)=\sum_{n=0}^{\infty} a_nx^{n+r}, $$ where \(a_n\) are the coefficients to be determined, and r is a constant to be found.

Substituting the Frobenius solution into the standard form differential equation

Let's calculate the derivatives of the Frobenius solution: First derivative: $$ y^\prime(x)=\sum_{n=0}^{\infty}(n+r)a_nx^{n+r-1}. $$ Second derivative: $$ y^{\prime\prime}(x)=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-2}. $$ Now, let's substitute the Frobenius solution and its derivatives into the standard form differential equation: $$ \sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n+r-2}+\left(\frac{4+7 x^{2}}{x\left(1+x^{2}\right)}\right) (\sum_{n=0}^{\infty}(n+r)a_nx^{n+r-1})+\left(\frac{8 x}{x\left(1+x^{2}\right)}\right)(\sum_{n=0}^{\infty} a_nx^{n+r})=0 $$

Developing the recurrence relation

Let's multiply each term in the equation by the corresponding powers of x: \( x^{r-2}\left[\sum_{n=0}^{\infty}(n+r)(n+r-1)a_nx^{n}\right]+x^{r-1}\left(\frac{4+7 x^{2}}{1+x^{2}}\right) \left[\sum_{n=0}^{\infty}(n+r)a_nx^{n}\right]+x^{r}\left(\frac{8}{1+x^{2}}\right)\left[\sum_{n=0}^{\infty} a_nx^{n}\right]=0 \) Comparing the coefficients of the powers of x, we obtain the following recurrence relation: $$ (n+r)(n+r-1)a_n+\left(\frac{4+7 (n+r-1)^2}{(n+r-1)(1+(n+r-1)^2)}\right)(n+r)a_{n-1}+\left(\frac{8}{1+(n+r-1)^2}\right) a_{n-2}=0, \quad \text{for} \quad n>1. $$

Finding explicit formulas for the coefficients in the Frobenius solutions

From the above recurrence relation, we can express \(a_n\) in terms of \(a_{n-1}\) and \(a_{n-2}\): $$ a_n=-\frac{(n+r)(n+r-1)}{(4+7(n+r-1)^2)(n+r-1)(1+(n+r-1)^2)}a_{n-1}-\frac{8}{(4+7(n+r-1)^2)(1+(n+r-1)^2)}a_{n-2}, \quad \text{for} \quad n>1. $$ To find explicit formulas for the coefficients in the Frobenius solutions, we need to calculate the first few terms based on the initial values, \(a_0\) and \(a_1\), and the value of r.

Presenting the Frobenius solutions

Finally, the fundamental set of Frobenius solutions can be presented as follows: $$ y(x)=c_1\left(a_0+\sum_{n=1}^{\infty} a_nx^{n+r}\right)+c_2\underbrace{\left(a_1+\sum_{n=1}^{\infty}\left[\frac{(n+r)(n+r-1)}{(4+7(n+r-1)^2)(n+r-1)(1+(n+r-1)^2)}\right] a_nx^{n+r}\right)}_{=b_1+a_{1,1}x^{r_1}+a_{2,1}x^{r_2}+\cdots}, $$ where \(c_1\) and \(c_2\) are constants, and the coefficients \(a_n\) are determined from the explicit formulas obtained in Step 5.

Key concepts

Differential Equations

Differential equations are at the heart of modeling many natural phenomena; they describe the relationship between a function and its derivatives. The equation from the exercise is a second-order linear homogeneous differential equation. These types of equations come up frequently in physical and engineering contexts, where the highest derivative is squared (second derivative in this case), and the solution is not multiplied by any term outside of the derivative—hence 'homogeneous.' In the context of Frobenius method, such equations can often require special series solutions when they have singular points, which ordinary power series cannot handle. This transition to the Frobenius method allows us to tackle differential equations that have power-law behaviors near singular points, common in physical situations like quantum mechanics or general relativity.

Power Series Solutions

Power series solutions are a potent tool for solving differential equations when no standard form can easily solve the equation. A power series is an infinite sum of the form \(\sum_{n=0}^{fty} a_nx^{n}\), where each term \(a_nx^n\) represents a coefficient \(a_n\) multiplied by the variable \(x\) raised to a power \(n\). When approaching differential equations with power series, we look for solutions that can be expressed as such series. The Frobenius method takes this idea one step further, considering solutions of the form \(y(x)=sum_{n=0}^{fty} a_nx^{n+r}\), introducing an additional parameter \(r\), which allows us to consider solutions around points where the standard power series might not converge. This is crucial for understanding where and why the Frobenius method is applied—particularly in equations with singular points where other methods aren't adequate.

Recurrence Relations

Recurrence relations express each term of a sequence as a function of its preceding terms. The Frobenius method invariably results in such relations, particularly when we substitute the assumed series form of the solution back into the differential equation. Recurrence relations provide a recipe for computing the coefficients of the series solution methodically. For instance, in the given exercise, knowing the initial conditions \(a_0\) and \(a_1\), and the parameter \(r\), allows us to utilize the established recurrence relation to find each subsequent coefficient \(a_n\). These recurrence relations are crucial as they enable us to construct the solution one coefficient at a time, thus building up the series solution as far as required for a satisfactory approximation or, in ideal cases, an exact solution.