Question

Find a fundamental set of Frobenius solutions. Give explicit formulas for the coefficients. $$ x^{2} y^{\prime \prime}+x\left(1+2 x^{2}\right) y^{\prime}-\left(1-10 x^{2}\right) y=0 $$

Short answer

Question: Find a fundamental set of Frobenius solutions for the given differential equation: \(x^2y''(x)+x(1+2x^2)y'(x)-(1-10x^2)y(x)=0\). Answer: The fundamental set of Frobenius solutions for the given differential equation is: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}, $$ where the coefficients \(a_n\) satisfy the recurrence relation: $$ a_{n+1} = \frac{(n+r)(n+r-1) a_n + (n-1+r)(n+r) a_{n-1} - (a_n - 10a_{n-2})}{2(n+1+r)(n+2+r)} $$ for \(n \geq 0\).

Step-by-step solution

Assume y as a Frobenius series

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

Derive y' and y''

Now calculate the derivatives of y(x): $$ y'(x) = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} $$ and $$ y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}. $$

Substitute y, y', and y'' into the given equation

Substitute y, y', and y'' from Steps 1 and 2 into the given differential equation: $$ x^{2} \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}+x\left(1+2 x^{2}\right) \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1}-\left(1-10 x^{2}\right) \sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$

Simplify and group terms

Now we need to simplify and group terms with the same power of x: $$ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r} + \sum_{n=0}^{\infty} (n+r) a_n x^{n+r+1} + 2\sum_{n=0}^{\infty} (n+r) a_n x^{n+r+3} - \left[\sum_{n=0}^{\infty} a_n x^{n+r} - 10\sum_{n=0}^{\infty} a_n x^{n+r+2}\right] = 0. $$

Equate coefficients

Since all coefficients with the same power of x should be equal, we can write: $$ (n+r)(n+r-1) a_n + (n-1+r)(n+r) a_{n-1} + 2(n+1+r)(n+2+r) a_{n+1} - (a_n - 10a_{n-2}) = 0, $$ for \(n \geq 0\).

Find the recurrence relation

Now we can find the recurrence relation for the coefficients: $$ a_{n+1} = \frac{(n+r)(n+r-1) a_n + (n-1+r)(n+r) a_{n-1} - (a_n - 10a_{n-2})}{2(n+1+r)(n+2+r)} $$ for \(n \geq 0\).

Generate the first few coefficients

Using the recurrence relation, we can generate the first few coefficients: $$ a_0, a_1, a_2, a_3, \dots $$

Conclusion

In conclusion, we have found a fundamental set of Frobenius solutions for the given differential equation: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r}. $$ where \(a_n\) is given by the following recurrence relation: $$ a_{n+1} = \frac{(n+r)(n+r-1) a_n + (n-1+r)(n+r) a_{n-1} - (a_n - 10a_{n-2})}{2(n+1+r)(n+2+r)} $$ for \(n \geq 0\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. In simpler terms, they tell us how a quantity changes in relation to its rate of change. These equations are widely used to model real-world phenomena in physics, engineering, biology, and more.
Understanding differential equations involves knowing how to identify the parts of the equation and the variables involved. In this context, our equation is:

• \( x^2 y'' + x(1 + 2x^2) y' - (1 - 10x^2)y = 0 \)

Here, \( y \) is the function we want to solve for, and \( y' \) and \( y'' \) are its first and second derivatives, respectively. The coefficients of these derivatives are functions of \( x \).
Solving differential equations often requires special methods, like the Frobenius method, when coefficients are not constant or when regular methods do not apply effectively.

Series Solutions

The series solution approach involves expressing the solution of the differential equation as an infinite series. This method is powerful, especially when dealing with complicated equations. The Frobenius method in particular expands solutions around a singular point by assuming a solution in the form of a power series:

• \( y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} \)

Here, \( a_n \) are the unknown coefficients to be determined, and \( r \) is a constant possibly representing the order of the singularity.
The advantage of this method is that it transforms the problem into calculating these coefficients, which can often be done systematically. By substituting the series into the original equation, one can match coefficients of like powers of \( x \) to get a series of recursive formulas.

Recurrence Relations

Recurrence relations are equations that express each term of a sequence as a function of preceding terms. In the series solution of differential equations, they play a vital role in determining the coefficients \( a_n \).
When we substituted the power series into our differential equation, we derived the recurrence relation:

• \( a_{n+1} = \frac{(n+r)(n+r-1) a_n + (n-1+r)(n+r) a_{n-1} - (a_n - 10a_{n-2})}{2(n+1+r)(n+2+r)} \)

This formula helps us calculate the next coefficient based on the previous ones. Solving these allows one to build up terms of the power series step by step, starting from an initial condition, often denoted \( a_0 \).
Recurrence relations simplify complex problems by breaking them into simpler repetitive tasks, as seen in recursive computer algorithms.

Coefficient Determination

To solve the differential equation using series solutions, determining the coefficients \( a_n \) is crucial. These coefficients define the solution series, and their calculation is guided by the recurrence relation.
Here's how it generally works:

• Start with initial values, such as \( a_0 \), which can sometimes be arbitrarily chosen unless specified by boundary conditions.
• Use the recurrence relation to determine subsequent coefficients.

By plugging into the relation:

• Set initial conditions based on physical, boundary, or exact problem requirements.
• Calculate \( a_1, a_2, \) and beyond, applying the relation step by step.

The calculated coefficients form the series solution, which approximates the actual function over a domain around the singular point. The beauty of this method lies in its systematic approach to breaking down complex problems, one coefficient at a time.