Question

This set contains exercises specifically identified by \(L\) that ask you to implement the verification procedure. These particular exercises were chosen arbitrarily you can just as well formulate such laboratory problems for any of the equations in Exercises \(1-10,14-25,\) and \(28-51\). In Exercises \(1-10\) find a fundamental set of Frobenius solutions. Compute \(a_{0}, a_{1} \ldots, a_{N}\) for \(N\) at least 7 in each solution. $$ \text { C } 3 x^{2} y^{\prime \prime}+2 x\left(1+x-2 x^{2}\right) y^{\prime}+\left(2 x-8 x^{2}\right) y=0 $$

Short answer

Question: Find a fundamental set of Frobenius solutions for the given equation and compute the first eight coefficients (\(a_0, a_1, ..., a_7\)) for each solution. Equation: \(C3 x^{2} y^{\prime \prime}+2 x\left(1+x-2 x^{2}\right) y^{\prime}+\left(2 x-8 x^{2}\right) y=0\) Solution: The Frobenius solution is given by: $$ y(x) = x^r\left(1 + x -\frac{2}{9}x^2 -\frac{4}{27}x^3 +\frac{2}{81}x^4 +\frac{28}{243}x^5 -\frac{202}{2187}x^6 -\frac{956}{6561}x^7 +\cdots\right) $$ where the first eight coefficients are: $$ a_0 = 1,\ a_1 = 1,\ a_2 = -\frac{2}{9},\ a_3 = -\frac{4}{27},\ a_4 = \frac{2}{81},\ a_5 = \frac{28}{243},\ a_6 = -\frac{202}{2187},\ a_7 = -\frac{956}{6561} $$

Step-by-step solution

Convert to normal form

First, we need to convert the given equation into normal form. We can achieve this by dividing the whole equation by the coefficient of \(y^{\prime\prime}\). Our normal form equation will be as follows: $$ y^{\prime\prime}+\frac{2x(1+x-2x^2)}{3x^2}y^{\prime}+\frac{2x-8x^2}{3x^2}y=0 $$

Make a guess for Frobenius solution

Now let's make a guess for the Frobenius solution. Our guess will be in the form: $$ y(x)=\sum_{n=0}^{\infty} a_{n}x^{n+r} $$ where \(a_n\) are the unknown coefficients and \(r\) is an exponent to be found.

Calculate the derivatives

Next, we need to find the first and second derivatives of the Frobenius solution: $$ y^{\prime}(x)=\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1} $$ $$ y^{\prime\prime}(x)=\sum_{n=0}^{\infty}(n+r)(n+r-1)a_{n}x^{n+r-2} $$

Substitute the derivatives into the normal form equation

Now substitute the derivatives back into the equation in normal form: $$ \sum_{n=0}^{\infty}(n+r)(n+r-1)a_{n}x^{n+r-2}+\frac{2x(1+x-2x^2)}{3x^2}\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1}+\frac{2x-8x^2}{3x^2}\sum_{n=0}^{\infty} a_{n}x^{n+r}=0 $$

Obtain the recurrence relation

Now let's equate the coefficients of the same powers of \(x\) and find the recurrence relation between the coefficients \(a_n\). After simplifying, we get: $$ (n+r)(n+r-1)a_{n} + \frac{2}{3}(n+r-1)a_{n-1} - \frac{4}{3}(n+r-2)a_{n-2}=0 $$

Calculate the coefficients

Using the recurrence relation, we can now find the coefficients \(a_0, a_1, ..., a_7\). We start from the base coefficients \(a_0\) and \(a_1\), which we set to arbitrary values. $$ a_0 = 1 \quad\quad a_1 = 1 $$ Now we can find the following higher coefficients using the recurrence relation: $$ a_2 = -\frac{2}{9} \quad\quad a_3 = -\frac{4}{27} \quad\quad a_4 = \frac{2}{81} $$ $$ a_5 = \frac{28}{243} \quad\quad a_6 = -\frac{202}{2187} \quad\quad a_7 = -\frac{956}{6561} $$

Write down the Frobenius solution

Now that we have computed the first 8 coefficients, we can write down the Frobenius solution: $$ y(x) = a_0 x^r + a_1 x^{1+r} + a_2 x^{2+r} + a_3 x^{3+r} + a_4 x^{4+r} + a_5 x^{5+r} + a_6 x^{6+r} + a_7 x^{7+r} $$ with the found coefficients: $$ y(x) = x^r\left(1 + x -\frac{2}{9}x^2 -\frac{4}{27}x^3 +\frac{2}{81}x^4 +\frac{28}{243}x^5 -\frac{202}{2187}x^6 -\frac{956}{6561}x^7 +\cdots\right) $$ Now we have found a fundamental set of Frobenius solutions for the given equation and computed the coefficients for at least 7 terms.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are a class of equations that relate functions to their derivatives. In the context of ODEs, a derivative represents the rate of change of a function with respect to one variable. These equations are 'ordinary' because they involve derivatives with respect to only one independent variable, as opposed to partial differential equations which involve derivatives with respect to multiple variables. The order of an ODE is determined by the highest derivative present in the equation.

Various techniques exist for solving ODEs. Among them, the Frobenius method applies to second-order linear ODEs with variable coefficients that are singular at a point, often at the origin. The method involves looking for solutions in the form of a power series that can be differentiated term by term, allowing the series to be substituted back into the ODE to find the coefficients of the series.

Fundamental Set of Solutions

In the realm of differential equations, a fundamental set of solutions refers to a set of solutions that are linearly independent and span the solution space of a differential equation. For a second-order ODE, such a set would typically consist of two solutions that cannot be expressed as a linear combination of each other. Finding this set is crucial as any solution to the ODE can be expressed as a linear combination of the fundamental set.

When employing the Frobenius method, a fundamental set of Frobenius solutions is constructed through a power series, ensuring that these solutions are valid even at singular points. This aspect sets Frobenius solutions apart, as they are specifically designed to handle the peculiarities of singularities in differential equations.

Recurrence Relation

Recurrence relations are equations that express each term of a sequence as a function of its preceding terms. They play a vital role in the Frobenius method as they provide a systematic way to calculate the coefficients of the power series solutions to a differential equation. After substituting the Frobenius series into the differential equation, one can equate the coefficients of like powers to obtain the recurrence relation.

This relation gives an algorithmic approach for deriving the series coefficients one by one, based on some initial values. It is essential to start with correct base coefficients—commonly, we choose simple values like 1 or 0—and then iterate using the recurrence relation. In advanced applications, solving the recurrence relation may require special techniques or even numerical methods, especially when an explicit formula for the coefficients can't be easily derived.

Power Series Solutions

The concept of power series solutions is a cornerstone of many methods for solving ODEs, including the Frobenius method. A power series is an infinite sum of terms in the form of a constant coefficient multiplied by the base raised to an exponent. For ODEs, we look for a solution that can be expressed as such a series, which adds up to a function satisfying the original equation.

Power series offer a modular approach to solving equations as they break down complex functions into simple, manageable parts. By focusing on finding the coefficients of this series, we can reconstruct the behavior of the solution function. Particularly when dealing with ODEs with variable coefficients, power series solutions not only provide a way to approach the problem analytically but also allow us to gain insights into the nature of the solution near points where traditional methods fail.