Question

Identify the singular point. Find the general solution that is valid for values of \(t\) on either side of the singular point. $$ t^{2} y^{\prime \prime}-t y^{\prime}+5 y=0 $$

Short answer

Answer: Frobenius method.

Step-by-step solution

Identify Singular Points

We will first classify the singular points by checking if any coefficients become infinite or not continuous. In the given ODE, we can rewrite the equation as: $$ y^{\prime \prime} - \frac{1}{t} y^{\prime} + \frac{5}{t^2} y = 0 $$ We see that coefficient \(\frac{1}{t}\) and \(\frac{5}{t^2}\) become infinite when \(t=0\). Thus, \(t=0\) is the singular point of the given equation.

Apply the Frobenius Method

To find the general solution around the singular point using the Frobenius method, we assume a solution in the following form: $$ y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} $$ where \(r\) is a constant and \(a_n\) are the coefficients to be determined.

Find Recursion Relation

Let's first determine the first and second derivatives with respect to \(t\): $$ y^{\prime}(t) = \sum_{n=0}^{\infty} (n+r) a_n t^{n+r-1} $$ $$ y^{\prime \prime}(t) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n t^{n+r-2} $$ Substitute the expressions for \(y(t)\), \(y^{\prime}(t)\), and \(y^{\prime \prime}(t)\) into the given ODE: $$ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n t^{n+r} - \sum_{n=0}^{\infty} (n+r) a_n t^{n+r} + 5\sum_{n=0}^{\infty} a_n t^{n+r} = 0 $$ To find the recurrence relation for coefficients \(a_n\), we need to make sure the powers of \(t\) are equal on each term. We first change all sum limits to start from \(n=1\) and shift \(n-1 \rightarrow n\) in the first term and \(n-2 \rightarrow n\) in the second term: $$ \sum_{n=1}^{\infty} (n+r+1)(n+r) a_{n-1} t^{n+r} - \sum_{n=1}^{\infty} (n+r) a_n t^{n+r} + 5\sum_{n=0}^{\infty} a_n t^{n+r} = 0 $$ Now all terms have the same power of \(t\), so we equate each term of the series by comparing the coefficients of \(t^{n+r}\): $$ (n+r+1)(n+r) a_{n-1} - (n+r) a_n + 5a_n = 0 $$ Solve for \(a_n\) to obtain the recurrence relation: $$ a_n = \frac{(n+r+1)(n+r)}{(n+r) - 5} a_{n-1} $$

Determine the General Solution

Now that our recursion relation is established, we proceed to determine the general solution. For the given ODE, the solution is a function \(y(t)\) that is valid on both sides of the singular point \(t=0\). The general solution that satisfies the recurrence relation can be expressed as: $$ y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} $$ The specific values of \(a_n\) and \(r\) will depend on the boundary conditions and initial conditions given for the problem. However, the general structure of the solution is established through the Frobenius method and the obtained recurrence relation.

Key concepts

Singular Point

When dealing with differential equations, a singular point is of particular importance. It is a value for the independent variable, usually denoted by t, where the coefficients of the differential equation are not defined or are discontinuous. Identifying singular points is the first step in many solution methods for differential equations.

In the context of the problem we're looking at, which is described by the differential equation \( t^{2} y'' - t y' + 5 y = 0 \), the singular point occurs at \( t = 0 \). This is because the coefficients \( \frac{1}{t} \) and \( \frac{5}{t^2} \) featured in the rearranged equation become undefined at that specific value of \( t \). Understanding where these singularities occur allows us to apply appropriate techniques, such as the Frobenius method, to find a series solution around this point.

General Solution of Differential Equation

The phrase general solution of a differential equation refers to a solution that encompasses all possible particular solutions to that equation. It typically includes arbitrary constants or functions, reflecting the fact that differential equations possess an infinite set of solutions corresponding to various initial or boundary conditions.

For linear second-order differential equations like the one in our exercise, the general solution usually consists of two linearly independent solutions combined together. When using the Frobenius method, we represent the general solution as an infinite series, \( y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} \), where \( a_n \) are the coefficients determined through a recursion formula, and \( r \) is a constant to be found. This series represents the general solution in the neighborhood of the singular point \( t=0 \) and incorporates all possibilities for behavior as \( t \) moves away from this singular point.

Recurrence Relation

In solving differential equations using series solutions, a recurrence relation is a formula that relates each term in the series to previous terms. It is fundamental in determining the coefficients of the series that represents the solution to the differential equation.

In the exercise, to establish the recurrence relation, derivatives of the proposed series solution are substituted back into the differential equation. The resulting expression is organized so that powers of \( t \) match, which enables us to equate coefficients of like powers of \( t \) and obtain a relationship between the coefficients \( a_n \) of the series. For instance, the derived relation \( a_n = \frac{(n+r+1)(n+r)}{(n+r) - 5} a_{n-1} \) tells us how to find any coefficient \( a_n \) from the preceding one \( a_{n-1} \) in the series. This recurrence relation is pivotal to building the general solution term by term, starting from an initial known coefficient, often \( a_0 \) or \( a_1 \) depending on the differential equation.