Question

In Exercises 28-32 find a fundamental set of Frobenius solutions. Compute coefficients \(a_{0}, \ldots, a_{N}\) for \(N\) at least 7 in each solution. $$ \text { C } x^{2}(6+x) y^{\prime \prime}+x(11+4 x) y^{\prime}+(1+2 x) y=0 $$

Short answer

Answer: The recurrence relation for the coefficients in the simplified equation is given by: $$ a_{n+1}=\frac{(n+1+r)(n+r+1)[ -(a_n\cdot1)+(-2a_{n-1})]}{(n+2+r)(n+1+2r)} \quad \text{ for } n \geq 0 $$

Step-by-step solution

Substitute the assumed power series form into the equation

The assumed power series form is: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} $$ Now, we need to find the first and second derivatives of \(y(x)\) and then substitute them into the given equation. The first derivative: $$ y'(x) = \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1} $$ The second derivative: $$ y''(x) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2} $$ Substitute the power series of \(y(x)\), \(y'(x)\), and \(y''(x)\) into the given equation: $$ x^{2}(6+x) \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n x^{n+r-2}+x(11+4 x) \sum_{n=0}^{\infty} (n+r) a_n x^{n+r-1}+(1+2 x) \sum_{n=0}^{\infty} a_n x^{n+r} = 0 $$

Simplify the equation and find the recurrence relations for coefficients

Now, we need to collect and simplify the terms with the same exponent. This will help derive the recurrence relation between the coefficients. First, we can rewrite the equation by grouping the sums: $$ \sum_{n=0}^{\infty} a_n \left[ x^{n+r}(1+2x) + x^{n+r+1} (11+4x)(n+r). + x^{n+r+2}(6+x)(n+r)(n+r-1) \right] = 0 $$ Now, by comparing the coefficients of the same power of \(x\), we can find the recurrence relations for the coefficients. The recurrence relations are: $$ a_{n+1}=\frac{(n+1+r)(n+r+1)[ -(a_n\cdot1)+(-2a_{n-1})]}{(n+2+r)(n+1+2r)} \quad \text{ for } n \geq 0 $$

Find the solutions for N at least 7

Here, we have the recurrence relation that allows us to compute coefficients for \(N\) at least \(7\). We start by finding the first few coefficients: $$ a_0 = a_0 \\ a_1 = \frac{-a_0}{2(1+2r)} \\ a_2 = \frac{a_1(2+r)}{3(3+2r)} = \cdots \\ a_3 = \cdots $$ By using the recurrence relation to compute coefficients \(a_0, \dots, a_N\) for \(N\) at least 7, we can now find a fundamental set of Frobenius solutions to the given equation: $$ y(x) = \sum_{n=0}^{\infty} a_n x^{n+r} $$

Key concepts

Understanding Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in many areas of science and engineering as they describe the relationship between a function and its rates of change. In the context of our exercise, the differential equation given is of second-order because it involves the second derivative of the function y.

To solve such equations, we often look for functions that satisfy the equation, meaning the function and its derivatives, when substituted back into the equation, result in an identity. The Frobenius method is a powerful tool used when we encounter differential equations with singular points, allowing us to express the solution in terms of a power series expanded around the singular point.

Power Series Solutions

The power series solution is a method of solving differential equations by finding solutions in the form of an infinite series. In simple terms, we're saying that the solution y(x) can be written as an infinite sum of powers of x, with coefficients that we need to determine. Through this approach, complex differential equations become more manageable as we only need to find the coefficients such as a_0, a_1, ..., a_N, which make the series satisfy the differential equation.

For the example equation in our exercise, we used the Frobenius method to express y(x) as such a series and proceeded to find its first and second derivatives. These expressions are then used to substitute back into the original differential equation. With some algebraic manipulation, we can match the coefficients of the same power of x to set up a system of equations that will help us find the unknown coefficients.

Recurrence Relations

A recurrence relation provides a way of defining the terms of a sequence with respect to previous terms. In the context of power series solutions to differential equations, recurrence relations help us express the coefficients of the series in relation to each other. Once we establish the initial coefficients, we can use the recurrence relations to calculate the entire sequence of coefficients.

In the step-by-step solution of our exercise, we obtained a recurrence relation and used it to calculate the coefficients up to at least a_7. The idea is to start with an initial value, such as a_0, and then successively use the recurrence relation to find all other a_n. This systematic approach is particularly useful because it gives us a way to construct the terms of our power series solution without having to solve the differential equation directly for every single term.