Question

Identify the singular point. Find the general solution that is valid for values of \(t\) on either side of the singular point. $$ y^{\prime \prime}+\frac{5}{t} y^{\prime}+\frac{40}{t^{2}} y=0 $$

Short answer

The singular point of the given differential equation is \(t=0\), and the general solution valid for values of \(t\) on either side of the singular point is \(y(t) = C_1 t^{-1}\sum_{n=0}^{\infty} a_n t^n + C_2 t^{-4}\sum_{n=0}^{\infty} a_n t^n\), where \(C_1\) and \(C_2\) are constants.

Step-by-step solution

Identify the singular points

We are given the differential equation: $$ y^{\prime \prime} + \frac{5}{t} y^{\prime} + \frac{40}{t^{2}} y = 0 $$ The singular points are the values of \(t\) that make the coefficients of the derivatives not analytic (in other words, infinite). We can see that only \(t=0\) is a singular point, as it makes the coefficients of \(y'\) and \(y\) infinite.

Analyzing the behavior near the singular point

Near the singular point \(t=0\), we can rewrite the given equation as $$ t^2 y^{\prime\prime} + 5t y^{\prime} + 40 y = 0 $$ Let's express the solution of the differential equation in terms of a series: $$ y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} $$ Where \(a_n\) are the coefficients of the series and \(r\) is an unknown exponent.

Obtain the general solution using Frobenius method

Differentiating the series expression for y(t) twice, we get: $$ y'(t) = \sum_{n=0}^{\infty} (n+r) a_n t^{n+r-1} \\ y^{\prime\prime}(t) = \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n t^{n+r-2} $$ Now let's substitute the expressions of \(y(t)\), \(y'(t)\), and \(y^{\prime\prime}(t)\) into the original differential equation: $$ t^2\sum_{n=0}^{\infty} (n+r)(n+r-1) a_n t^{n+r-2} + 5t\sum_{n=0}^{\infty} (n+r) a_n t^{n+r-1} + 40\sum_{n=0}^{\infty} a_n t^{n+r} = 0 $$ In order to find the coefficients \(a_n\) recursively and the exponent \(r\), we need to simplify and balance the powers of \(t\) on both sides of the equation. We can start by canceling out the powers of \(t\) within the sums: $$ \sum_{n=0}^{\infty} (n+r)(n+r-1) a_n t^{n+r} + 5\sum_{n=0}^{\infty} (n+r) a_n t^{n+r} + 40\sum_{n=0}^{\infty} a_n t^{n+r} = 0 $$ Now, we can combine the sums: $$ \sum_{n=0}^{\infty} \left[(n+r)(n+r-1)a_n + 5(n+r)a_n + 40a_n\right] t^{n+r} = 0 $$ Since the series is equal to zero for every power of \(t\), we must have: $$(n+r)(n+r-1) a_n + 5(n+r) a_n + 40 a_n = 0$$ Now, we can use the recurrence relation to find \(a_n\) for \(n \geq 1\): $$ a_n = -\frac{5(n+r)+40}{(n+r)(n+r-1)} a_{n-1} $$ To find the exponent \(r\), we must examine the indicial equation (when \(n=0\)): $$ r(r-1) a_0 + 5r a_0 + 40 a_0 = 0 $$ Notice that \(a_0 \neq 0\) as it needs to be a non-trivial solution. From this, we find \(r(r-1) + 5r + 40 = (r+4)(r+1) = 0\) with roots \(r_1 = -1\) and \(r_2 = -4\). So, there are two linearly independent solutions: 1. \(y_1 (t) = t^{-1}\sum_{n=0}^{\infty} a_n t^n\) 2. \(y_2(t) = t^{-4}\sum_{n=0}^{\infty} a_n t^n\) Now we have found two linearly independent solutions, implying the general solution is their linear combination: $$ y(t) = C_1 t^{-1}\sum_{n=0}^{\infty} a_n t^n + C_2 t^{-4}\sum_{n=0}^{\infty} a_n t^n $$

Final Answer

The singular point of the given differential equation is \(t=0\). The general solution that is valid for values of \(t\) on either side of the singular point is: $$ y(t) = C_1 t^{-1}\sum_{n=0}^{\infty} a_n t^n + C_2 t^{-4}\sum_{n=0}^{\infty} a_n t^n $$

Key concepts

Singular Point

In differential equations, a singular point is a value where the coefficients of the equation become undefined or infinite. For the given differential equation, \( y'' + \frac{5}{t}y' + \frac{40}{t^2}y = 0 \), the singular point is \( t = 0 \). This happens because at \( t = 0 \), the coefficients \( \frac{5}{t} \) and \( \frac{40}{t^2} \) become infinite. Detecting singular points is crucial because they can change the behavior of solutions near those points. Understanding where a singular point occurs helps in determining whether solutions can be extended over the singularity or need special methods to solve around it. When dealing with such singularities, the Frobenius method often comes into play.

Frobenius Method

The Frobenius Method is a powerful tool used to find series solutions to differential equations at singular points, specifically at regular singular points. It generalizes the power series method by allowing solutions of the form \( y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} \), where \( r \) is not necessarily an integer.

• Start with an assumed series: \( y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} \).
• Differentiate the series to find expressions for \( y'(t) \) and \( y''(t) \).
• Substitute these into the differential equation to determine a recursion relation for the coefficients \( a_n \).

The main goal is to determine the unknown exponent \( r \) and the coefficients \( a_n \). The ability of the method to accommodate non-integer exponents makes it flexible for solving a wide range of problems involving singular points.

Series Solution

A series solution is an expression of the solution to a differential equation as a sum of terms in a series. Using the Frobenius method, we assume a solution of the form \( y(t) = \sum_{n=0}^{\infty} a_n t^{n+r} \). This technique is particularly useful when dealing with differential equations that cannot be solved through simple algebraic methods.

• Series solutions are often used near singular points to represent complex functions in a manageable format.
• The coefficients \( a_n \) in the series need to satisfy a recurrence relation derived from substituting the series back into the differential equation.
• These coefficients can often be determined recursively, starting with an initial condition or the smallest term.

The series approach allows for solutions that might not be expressible in terms of standard functions, providing significant insight into the behavior of solutions across different domains.

Indicial Equation

The indicial equation is a crucial part of applying the Frobenius method. It is derived from the lowest power of \( t \) in the series solutions after substituting into the differential equation. The roots of the indicial equation determine the possible values of the exponent \( r \).

• The indicial equation often takes the form \( r(r-1) + ar + b = 0 \).
• Solving this quadratic equation provides two roots, which correspond to different possible series solutions.
• These roots highlight the basic structure and behavior of the series solution near the singular point.

For the example provided, the indicial equation results in \( (r+4)(r+1) = 0 \) with roots \( r_1 = -1 \) and \( r_2 = -4 \). These values illustrate that the differential equation possesses two linearly independent solutions, each corresponding to one of these roots. Understanding the indicial equation is essential to finding the correct form of the series and ensuring the completeness of solutions around singular points.