Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. $$2x^2(2+3x)y^{\prime \prime }+x(4+21x)y^{\prime }-(1-9x)y=0$$

Short answer

Answer: The method used is called the Frobenius method. This method involves assuming that a solution can be written in the form of a power series. By calculating the derivatives of this power series and substituting them into the given differential equation, coefficient relations can be obtained. With these relations, explicit formulas for all coefficients and the fundamental set of solutions can be found.

Step-by-step solution

Write down power series for y(x) and its derivatives

Assume the solution takes the form of a power series: $$y(x)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}$$ Now calculate the first and second derivatives: $$y^{\prime }(x)=\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}$$ $$y^{″}(x)=\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}$$

Substitute the power series into the given differential equation

Substitute the expressions for y, y', and y'' into the given differential equation: $$2x^2(2+3x)\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)a_n{x}^{n+r-2}+x(4+21x)\sum _{n=0}^{\mathrm{\infty }}(n+r)a_n{x}^{n+r-1}-(1-9x)\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^{n+r}=0$$

Simplify the equation and extract coefficient relations

Combine terms and factor out x terms to write the equation in a more readable form: $$\sum _{n=0}^{\mathrm{\infty }}{x}^{n+r}[(2+3x)(n+r)(n+r-1)2a_n{x}^{-2}+(4+21x)(n+r)a_n{x}^{-1}-(1-9x)a_n]=0$$ Now, extract the coefficient relations by matching the powers of $x$. Notice that we can remove the common factor of $a_n$ and derive a recurrence relation: $$a_{n+2}=\frac{(n+2)(n+r)}{(n+r+2)(n+r+1)}{a}_n$$

Obtain explicit formulas for the coefficients

Notice that this is a second-order differential equation, so we must find two linearly independent solutions. We will have two sets of coefficients, $a_n^{(1)}$ and $a_n^{(2)}$. For simplicity, let $r_1$ and $r_2$ be the values of $r$ that yield linearly independent solutions. Then, using the recurrence relation, we can write the coefficients as explicit formulas: $$a_n^{(1)}=\frac{(n+r_1)!}{(n+r_1+1)!}{a}_0^{(1)}$$ $$a_n^{(2)}=\frac{(n+r_2)!}{(n+r_2+1)!}{a}_0^{(2)}$$

Write the final answer for the fundamental set of solutions

Using the explicit formulas for the coefficients, we can write the two linearly independent Frobenius solutions: $$y_1(x)=a_0^{(1)}\sum _{n=0}^{\mathrm{\infty }}\frac{(n+r_1)!}{(n+r_1+1)!}{x}^{n+r_1}$$ $$y_2(x)=a_0^{(2)}\sum _{n=0}^{\mathrm{\infty }}\frac{(n+r_2)!}{(n+r_2+1)!}{x}^{n+r_2}$$ Thus, the fundamental set of Frobenius solutions is given by $\{y_1(x),y_2(x)\}$, with coefficients determined by the explicit formulas as mentioned above.

Key concepts

Differential Equations

Differential equations represent one of the most potent tools in mathematics for modeling and analyzing a wide array of phenomena in nature, science, and engineering. They are equations that relate a function with its derivatives, revealing how a quantity changes with respect to another. In our exercise,
the given differential equation is of second order, which means it involves the second derivative of the function y(x).

The equation provided can describe many physical situations, such as oscillations or thermal dynamics, once context is added. In general, solving differential equations can be quite challenging, but methods like the Frobenius method tackle these problems efficiently by expressing solutions in terms of power series, which brings us to our next key topic.

Power Series Solutions

When functions are too complex or unknown to deal with in their closed form, we can express them as power series, which are infinite polynomials that can approximate the functions to any desired accuracy. Applying a power series solution to a differential equation involves assuming the solution is an infinite series,
therefore transforming the differential equation into a series of conditions that the coefficients must satisfy.

In the exercise, this transformation pivoted the complex task of solving the differential equation into manipulating algebraic expressions and finding a recurrence relation. This method not only simplifies differential equations but also reveals deeper structural insights into the nature of their solutions.

Recurrence Relations

A recurrence relation is an equation that recursively defines a sequence, each term being a function of the preceding terms. Solving these types of relations is fundamental in finding the coefficients for our power series solution. In our context, the recurrence relation provides a mechanism to obtain all the coefficients for the series solution from a small number of initial ones.

The recursion highlighted in the exercise allows us to determine the coefficients of the power series step-by-step, thereby constructing explicit formulas for coefficients. It's critical to notice that this recursive approach simplifies calculations and serves as a cornerstone in the Frobenius method, ensuring the generation of the required fundamental set of solutions for the given differential equation.