Question

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

Short answer

Question: Find a fundamental set of Frobenius solutions for the given differential equation: $$ x^2 y'' - 3x y' + (3+4x) y = 0 $$ Answer: To find a fundamental set of Frobenius solutions, follow these steps: 1. Rewrite the differential equation in standard form. 2. Assume a solution in the form of a power series. 3. Calculate the first and second derivatives. 4. Substitute the derivatives into the standard form of the equation. 5. Simplify the equation by removing the common factors. 6. Apply the method of equating the coefficients. 7. Solve for \(a_n\). 8. Find the values of \(r\) for which the power series converges. 9. Substitute the values of \(r\) for the convergent power series. After following these steps, we get the recurrence relation for the coefficients of the power series as: $$ a_n = -\frac{4 a_{n-1}}{(n+r)(n+r-1) - 3(n+r) + 3} $$ The indicial equation is obtained by substituting \(n=0\): $$ r(r-1) - 3r + 3 = 0 $$ Solve this quadratic equation to obtain the possible values of \(r\). Finally, use these values to form different convergent power series, and use the recurrence relation to find the coefficients \(a_n\). These series form the fundamental set of Frobenius solutions.

Step-by-step solution

Rewrite the differential equation in standard form

Divide the entire equation by \(x^2\) to rewrite the equation in standard form: $$ y'' - \frac{3}{x} y' + \frac{3+4x}{x^2} y = 0 $$

Assume a solution in the form of a power series

Assume a solution of the form: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} $$ where \(a_n\) are the coefficients and \(r\) is an unknown number.

Calculate the first and second derivatives

Differentiate \(y(x)\) with respect to \(x\) twice to obtain: $$ y'(x) = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} $$ $$ y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} $$

Substitute the derivatives into the standard form of the equation

Plug \(y, y'\), and \(y''\) into the rewritten equation: $$ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} - \frac{3}{x} \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} + \frac{3+4x}{x^2} \sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$

Simplify the equation by removing the common factors

First, multiply the second and third summation by \(x^{2}\) and \(x\), respectively, then equate the powers of \(x\). $$ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r} - 3 \sum_{n=0}^{\infty} (n+r) a_n x^{n+r} + (3+4x) \sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$

Apply the method of equating the coefficients

In order for the equation to be satisfied for all \(x\), the coefficients of \(x^{n+r}\) in each term must sum up to zero. $$ ((n+r)(n+r-1) - 3(n+r) + 3) a_n + 4 a_{n-1} = 0 $$

Solve for \(a_n\)

Rearrange the equation to get an explicit formula for \(a_n\): $$ a_n = -\frac{4 a_{n-1}}{(n+r)(n+r-1) - 3(n+r) + 3} $$ This is the recurrence relation for the coefficients of the power series.

Find the values of \(r\) for which the power series converges

The indicial equation is obtained by substituting \(n=0\): $$ r(r-1) - 3r + 3 = 0 $$ Solve this quadratic equation to obtain the possible values of \(r\).

Substitute the values of \(r\) for the convergent power series

Use the obtained values of \(r\) to form different convergent power series. These series form the fundamental set of Frobenius solutions. Use the recurrence relation to find the coefficients \(a_n\).

Key concepts

Differential equations

In mathematics, differential equations are equations that involve an unknown function and its derivatives. They are powerful tools used to describe various phenomena, from the motion of planets to the flow of electricity. The specific type of differential equation presented in the exercise is known as a 'second-order linear homogeneous differential equation'.

These equations have the general form:
\[ a(x)y'' + b(x)y' + c(x)y = 0 \]
where \( y'' \) denotes the second derivative of the unknown function \( y \), and \( a(x) \), \( b(x) \), and \( c(x) \) are functions of \( x \). In the given problem, the differential equation
\[ x^{2} y'' - 3xy' + (3+4x)y = 0 \]
represents a situation where the coefficients of \( y \) and its derivatives are functions of \( x \). To solve such equations, various methods can be applied, one of which is the Frobenius method, particularly suitable for equations with singular points, as in the example provided.

Power series solutions

The Frobenius method relies on power series solutions to solve differential equations. A power series is an infinite sum of terms like \( a_n x^n \), where \( a_n \) is the coefficient of the nth term and can be thought of as a generalization of polynomials to an infinite number of terms. The assumption behind this method is that a solution can be expressed as such a series multiplied by a power of \( x \), that is to say,
\[ y(x) = \textstyle\bigg(\sum_{n=0}^{\infty} a_n x^{n}\textstyle\bigg) x^r \]
In the exercise, this is applied by assuming a solution in the form of \( y(x) = \textstyle\bigg(\sum_{n=0}^{\infty} a_n x^{n+r}\textstyle\bigg) \), with \( a_n \) representing the coefficients to be determined, and \( r \) being an unknown exponent that will be later found using the indicial equation. The goal is to turn the complex problem of solving a differential equation into determining the coefficients \( a_n \), which are simpler to handle.

Indicial equation

To find the exponent \( r \) in the power series method, we use the indicial equation. This is a key step in the Frobenius method and is derived by examining the lowest power of \( x \) (usually at \( n = 0 \)) in the power series.

In the example provided, by observing the coefficients in front of the lowest power of \( x \), we can construct an indicial equation:
\[ r(r-1) - 3r + 3 = 0 \]
This quadratic equation is formed after expressing the differential equation in terms of power series and considering the coefficients of the series where \( n = 0 \). Solving the indicial equation provides us with the values of \( r \), which are crucial in finding the complete solution to the differential equation as they dictate the behavior of the power series near the singular point, which is \( x = 0 \) in our case.

Recurrence relation

Once the indicial equation is solved, the recurrence relation is used to determine the coefficients \( a_n \) of the power series. This relation gives us a formula to compute each coefficient from the previous ones. The recurrence relation in our example is:
\[ a_n = -\frac{4 a_{n-1}}{{(n+r)(n+r-1) - 3(n+r) + 3}} \]
This provides a systematic way to find all coefficients, starting from the known initial one, to construct the full solution. It's like a mathematical domino effect; once you have the first piece, the subsequent pieces fall into place accordingly. Computing these coefficients might seem daunting, but it transforms an opaque differential equation into clear, manageable pieces, like solving a puzzle piece by piece. In the textbook exercise, correctly applying this recurrence relation and using the determined values for \( r \), students can establish the full power series solution to the original differential equation.