Question

Use the method of Frobenius to find solutions near \(x=0\) of each of the differential equations in Exercises, $$ x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-\frac{1}{9}\right) y=0 $$

Short answer

Using the method of Frobenius, we first rewrite the given differential equation with the basic Frobenius form and differentiate it twice with respect to x. We then substitute these derivatives back into the original differential equation and simplify. After matching powers of x in the series to create a recurrence relation between the coefficients, we find the value of r and a recursive relation for the coefficients. Finally, we plug the root values of r and the recursive relation for coefficients back into the basic Frobenius form: \(y(x) = x^r \sum \limits_{n = 0}^{\infty} a_nx^n\)

Step-by-step solution

(Step 1: Rewrite the given differential equation with basic Frobenius form)

To apply the method of Frobenius, rewrite the given differential equation with \(y = x^r \sum \limits_{n = 0}^{\infty} a_nx^n\). This will allow us to determine the coefficients of the power series.

(Step 2: Differentiate the basic Frobenius form)

Differentiate the basic Frobenius form twice with respect to \(x\): \(y = x^r\sum \limits_{n = 0}^\infty a_n x^n\) \(y' = r x^{r-1} \sum \limits_{n = 0}^\infty a_n x^n +x^r \sum \limits_{n = 1}^\infty n a_n x^{n-1} \) \(y'' = r(r-1) x^{r-2}\sum \limits_{n = 0}^\infty a_n x^n +2rx^{r-1} \sum \limits_{n = 1}^\infty n a_n x^{n-1} + x^r\sum \limits_{n = 2}^\infty n(n-1) a_n x^{n-2}\)

(Step 3: Substitute the derivatives into the original differential equation)

Substitute the derivatives back into the original differential equation and simplify: LHS: \(x^2y'' + xy' + (x^2 - \frac{1}{9})y\) = \(x^2[r(r-1) x^{r-2}\sum \limits_{n = 0}^\infty a_n x^n +2rx^{r-1} \sum \limits_{n = 1}^\infty n a_n x^{n-1} + x^r\sum \limits_{n = 2}^\infty n(n-1) a_n x^{n-2}] \) + \(x[r x^{r-1}\sum \limits_{n = 0}^\infty a_n x^n +x^r \sum \limits_{n = 1}^\infty n a_n x^{n-1}] \) + \([x^2 - \frac{1}{9})(x^r\sum \limits_{n = 0}^\infty a_n x^n)\] RHS = 0

(Step 4: Simplify the equation)

Simplify the equation and group terms. \(x^{r}\left[\sum \limits_{n = 0}^\infty a_n (r(r-1)x^n + 2rx^{n+1}n + x^{n+2}n(n-1))+ rx^{n+1}n + x^{n+2}n \\+ (x^n - \frac{1}{9}x^n)\right] = 0\)

(Step 5: Match powers of x)

Match powers of \(x\) in the series to create a recurrence relation between the coefficients: For \(n = 0\), we have: \(r(r-1)a_0 = 0\) For \(n=1\), we have: \(a_1(r(1)+r(r-1)+1) = 0 \) For \(n \geq 2\), we have: \(a_n(n(r+1)+r(r-1)+n(n-1))- \frac{1}{9}a_n = 0 \) After finding the value of r, plug the root values of r and a recursive relation for coefficients back into the basic Frobenius form: \(y(x) = x^r \sum \limits_{n = 0}^{\infty} a_nx^n\)

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. They describe various phenomena such as motion, growth, and decay. In the context of solving these equations, we aim to find a function or a set of functions that satisfy the given equation.

There are different types of differential equations:

• Ordinary Differential Equations (ODEs), which involve functions of a single variable and their derivatives.
• Partial Differential Equations (PDEs), which involve multiple variables and their partial derivatives.

The problem in the original exercise deals with an ordinary differential equation that is second-order. This means it involves the second derivative of a function.

The Frobenius method helps solve such ODEs specifically at points called singular points where solutions might not be as simple as they are at ordinary points.

Power Series Solutions

Power series solutions are a way to express the solution of a differential equation as an infinite sum. The general form of a power series solution is \( y(x) = \sum_{n=0}^{\infty} a_n x^n \).

This method is particularly useful for equations that cannot be solved by elementary functions like polynomials or exponential functions. In the context of the Frobenius method, the power series is adjusted slightly to include a factor of \(x^r\). Therefore, solutions take the form of: \( y(x) = x^r \sum_{n=0}^{\infty} a_n x^n \).

This adjustment allows the method to handle singular points by accommodating initial terms that may behave differently. By substituting back into the differential equation, these coefficients \(a_n\) can be determined, yielding the full series solution.

Recurrence Relations

Recurrence relations form the backbone of solving differential equations by the Frobenius method. Once the power series is substituted back into the differential equation and matched by powers of \(x\), it provides equations between the coefficients.

These equations are called recurrence relations. They describe how each coefficient relates to its neighbors, thus constructing the full series solution step by step:

• The base terms are identified first, often requiring special conditions like setting specific coefficients or recognizing initial values like \(a_0\), the leading term.
• The subsequent coefficients are calculated based on these base terms, allowing for iterative calculation of the series.

For example, after simplifying and matching powers of x, the exercise yields specific conditions for \(a_0, a_1\), and general conditions for subsequent \(a_n\). This ensures that all coefficients adhere to the equation's constraints, allowing us to find the solution using a step-by-step approach.

Ordinary Points

Ordinary points are specific values of \(x\) where a differential equation can be expressed in a simplified standard form, and solutions behave 'normally'. For an equation involving functions like \(x^2\) in its highest derivative term, an ordinary point is where all terms remain well-behaved — meaning no term of the form \(x^{-1}\).

In the context of the exercise, the Frobenius method allows for finding solutions at singular points near \(x = 0\), but these solutions also provide insights into behavior around ordinary points.

By converting the equation into a power series at the ordinary point or slightly perturbed due to a singularity, we can still analyze and predict how solutions behave. These insights are key in plotting the solution's behavior across a range of values, establishing an initial understanding of how the function evolves.