Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. \(9 x^{2} y^{\prime \prime}-3 x\left(7-2 x^{2}\right) y^{\prime}+\left(25+2 x^{2}\right) y=0\)

Short answer

In conclusion, to solve the given ODE, we applied the Frobenius method. We identified the regular singular point and indices, derived the indicial equation, and found the recurrence relations for the coefficients of the Frobenius series solutions. The fundamental set of Frobenius solutions can be expressed as: $$ y_1(x)=a_0 \left[ 1 -\frac{5}{26}x + \sum_{n=2}^{\infty} a_n x^n \right] $$ $$ y_2(x)=x \left[ a_0 + \sum_{n=1}^{\infty} a_{n+1} x^n \right] $$ These solutions are represented by the power series with coefficients found using the recurrence relations for each index, providing a general framework for solving this ODE.

Step-by-step solution

Identify the singular point

The given ODE can be written as $$ x^2(1+x^2)y''-x(5-x^2)y'+(7+25x^2)y = 0 $$ Divide the entire ODE by \(x^2\), we get $$ (1+x^{2})y^{\prime \prime}-(5x^{-1}-x)y^{\prime}+(7x^{-2}+25)y=0 $$ Since the equation is of the form \((1+x^2)y''-px'y'+qy=0\) with \(p=5x^{-1}-x\), and \(q=7x^{-2}+25\), we notice that the singular point of this ODE is at \(x=0\). It is a regular singular point because both \(p\) and \(q\) are analytic functions at \(x=0\).

Find the indicial equation

Now, let's try the series solution \( y = \sum_{n=0}^{\infty} a_n x^{n+r} \), with \(\displaystyle{a_0 \neq 0}\). The derivative and second derivative of the series is given by $$ y' = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} $$ $$ y'' = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} $$ Substituting the series solution and its derivatives into the ODE, we obtain the following: $$ (1+x^2) \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} - (5x^{-1} - x) \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} + (7x^{-2}+25 )\sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$ First let \(x=0\) to find the indicial equation. Collect the coefficients of \(x^r\) terms, we get $$ (r)(r-1) a_0 = 0 $$ This is the indicial equation. Since we know that \(a_0 \neq 0\), we have \(r(r-1)=0\), which gives \(r_1=0\) and \(r_2=1\) as the indices.

Find the recurrence relation

Going back to the series expression of the ODE, find the recurrence relation for each \(r\). For \(r=0 : x^{n}\), For \(n \geq 2\), the coefficients of \(x^n\) terms are: $$ (n)(n-1)a_n - 5(n-1)a_{n-1} + (7)a_{n-2} + 25 a_n = 0 $$ Rearrange to find a formula for \(a_n\): $$ a_n = \frac{5(n-1)a_{n-1} - 7 a_{n-2}}{(n-1)(n-26)} $$ For \(r=1 : x^{n+1}\), For \(n \geq 1\), the coefficients of \(x^{n+1}\) terms are: $$ (n+1)(n)a_{n+1} - 5na_n + (7)a_{n-1} + 25 a_{n+1} = 0 $$ Rearrange to find a formula for \(a_{n+1}\): $$ a_{n+1} = \frac{5na_n - 7 a_{n-1}}{(n)(n+25)} $$ Now, for each index, we have a recurrence relation that can be used to find the coefficients of the solution.

Find the coefficients

We will use the recurrence relations to find the first few coefficients of the solution. For each index, calculate the first few coefficients: For \(r_1=0\): \(a_0\) is arbitrary (non-zero), \( a_1 = \frac{5(-1)a_0}{(-1)(-26)}= -\frac{5}{26}a_0, a_2, a_3, ...\) For \(r_2=1\): \(a_0\) is arbitrary (non-zero), \(a_1, a_2, a_3, ...\) Now having calculated the coefficients, the fundamental set of Frobenius solutions for this equation is given by: $$ y_1(x)=a_0 \left[ 1 -\frac{5}{26}x + \sum_{n=2}^{\infty} a_n x^n \right] $$ $$ y_2(x)=x \left[ a_0 + \sum_{n=1}^{\infty} a_{n+1} x^n \right] $$ where the \(a_n\) are calculated using the recurrence relations.

Key concepts

Ordinary Differential Equations (ODEs)

Ordinary Differential Equations (ODEs) involve functions and their derivatives. They are crucial in modeling dynamic systems where variables change over time. An ODE like the one given describes how a function y depends on the variable x and its derivatives, such as \(y', y''\).

The primary job when working with ODEs is to find the function y that satisfies these relationships. Solutions provide insights into the behavior of physical, biological, or technological systems. In this exercise, the ODE is a second-order linear equation, meaning the highest derivative is second-order, making it slightly more complex. Understanding the structure of ODEs is essential for solving them using various methods like the Frobenius Method.

Singular Points

Singular points in differential equations are values where the solution might behave irregularly. They are points where the equation's coefficients have certain peculiar features. In our problem, the singular point is at \(x = 0\).

Identifying singular points is crucial because it informs how we should approach the solution.

• Regular Singular Point: The coefficients of the ODE like \(p(x)\) and \(q(x)\) are analytic at the singular point, meaning they behave nicely and are expressible as Taylor series.
• Irregular Singular Point: The coefficients exhibit more complicated behavior and require advanced methods for solutions.

Recognizing whether a singular point is regular or irregular determines the choice of solution methods, such as series solutions or the Frobenius Method.

Series Solutions

Series solutions involve expressing the solution of a differential equation as a power series. This is especially useful near singular points. Here, the series form \( y = \sum_{n=0}^{\infty} a_n x^{n+r} \) is used.

The power series allows us to find a solution even when standard methods fail. We assume a solution in terms of an infinite sum and then determine the coefficients \(a_n\) using recurrence relations. This method is very powerful for solving ODEs near singular points because it provides a way to capture the behavior of solutions in those regions. Once the series form is established, calculations reveal whether the series converges to a valid solution.

Recurrence Relations

Recurrence relations are used to find coefficients in series solutions. They provide a systematic way to calculate each coefficient based on previous ones.

For the given ODE, two recurrence relations were derived for different cases (\(r = 0\) and \(r = 1\)). These relations look like:

• \(r=0:\)
• \(a_n = \frac{5(n-1)a_{n-1} - 7 a_{n-2}}{(n-1)(n-26)}\)
• \(r=1:\)
• \(a_{n+1} = \frac{5na_n - 7 a_{n-1}}{(n)(n+25)}\)

By plugging in initial conditions or starting values, these formulas allow us to compute all necessary coefficients \(a_n\). The derivation and solution of these relations are key in constructing the series solution that addresses the behavior near the singular point. Understanding and applying them provide insight into the depth of a problem's solution.