Question

In each exercise, (a) Determine all singular points of the given differential equation and classify them as regular or irregular singular points. (b) At each regular singular point, determine the indicial equation and the exponents at the singularity. $$ \left(t^{2}-4\right)^{2} y^{\prime \prime}-y^{\prime}+y=0 $$

Short answer

Question: Identify the singular points of the given differential equation, classify them as regular or irregular, and find the indicial equation and exponents at each regular singular point for the following equation: $$(t^2-4)^2 y'' - (t^2-4)y' +y=0$$ Answer: The given differential equation has two regular singular points at t = 2 and t = -2. The indicial equation for both points is $$r^2 = 0$$. The exponents at the singular points are r = 0 and r = 1.

Step-by-step solution

Identify the Singular Points

First, we need to find the singular points, which are points at which the coefficients of the derivative terms in the equation become infinite. In this case, there are two points when the coefficient of the second derivative term becomes infinite: t = 2 and t = -2. Since the other coefficients don't depend on t, these two points will be the singular points. For a singular point to be considered regular, the equation must be of the form: $$ (t-t_0)^k y'' + p(t)(t-t_0)^{k-1} y' + q(t)(t-t_0)^{k-2} y = 0 $$ where p(t) and q(t) are analytic at $$t_0$$ and k is a positive integer. Our given differential equation can be written as: $$ (t^2-4)^2 y'' - (t^2-4)y' +y=0 $$ Now, comparing it to the above general form, it is clear that k for both singular points is 2, and that the equation is indeed regular at these singular points since the other coefficients (1 and -1) are constants and thus analytic at both of the singular points. Therefore, t = 2 and t = -2 are both regular singular points.

Determine the Indicial Equation for each Regular Singular Point

Now that we've identified the singular points, we need to find the indicial equation for the regular singular points. The indicial equation is given by the roots of the characteristic polynomial associated with the differential equation. To find the indicial equation, we assume a solution of the form: $$ y(t) = \sum_{n=0}^\infty a_n(t-t_0)^{n+r} $$ Now we substitute the assumed solution into our differential equation and then equate the coefficients of the various powers of \((t-t_0)\). We start with t = 2: $$ y(t) = \sum_{n=0}^\infty a_n(t-2)^{n+r} $$ After substituting to the differential equation and simplifying, we obtain the indicial equation for the regular singular point t = 2: $$ r(r-1) + r = 0 $$ Next, we'll repeat the process for t = -2: $$ y(t) = \sum_{n=0}^\infty a_n(t+2)^{n+r} $$ After substituting to the differential equation and simplifying, we obtain the indicial equation for the regular singular point t = -2: $$ r(r-1) + r = 0 $$ It is worth noting that in this case, both singular points have the same indicial equation.

Find the Exponents at each Regular Singular Point

Finally, we need to find the exponents at each regular singular point. We do this by finding the roots of the indicial equation. Since both singular points have the same indicial equation, we'll only need to find the roots once. We solve the indicial equation for r: $$ r^2 - r + r = r^2 = 0 $$ And we get the roots r = 0 and r = 1. Thus, we have found all necessary information for the given differential equation: 1. The singular points are t = 2 and t = -2, both of which are regular singular points. 2. The indicial equation for both points is $$r^2 = 0$$. 3. The exponents at the singularity for both points are r = 0 and r = 1.

Key concepts

Regular Singular Points

When dealing with differential equations, identifying singular points is crucial as they can significantly impact the behavior of solutions. Singular points are where the coefficients of the derivative terms become infinite. In our given equation, these occur at \( t = 2 \) and \( t = -2 \). To classify these as regular or irregular singular points, we need to compare the equation to the standard form:

• \[(t-t_0)^k y'' + p(t)(t-t_0)^{k-1} y' + q(t)(t-t_0)^{k-2} y = 0\]

where \( p(t) \) and \( q(t) \) are analytic at \( t_0 \), and \( k \) is a positive integer. For both \( t = 2 \) and \( t = -2 \), the rewritten equation aligns perfectly with this form with \( k = 2 \), and both points are consequently regular.
Since other coefficients, which are constants, remain analytic at these points, we conclude that both singularities are indeed regular. Regular singular points simplify the solution process as they allow application of specific techniques like the Frobenius method for solving differential equations.

Indicial Equation

The indicial equation is a crucial tool when solving differential equations at regular singular points. It arises from substituting a power series solution into the differential equation. The indicial equation helps determine the possible values of \( r \), which are vital for understanding the nature of solutions at each singularity.
For our equation, both singular points \( t = 2 \) and \( t = -2 \) yield the same indicial equation:

• \[r(r-1) + r = 0\]

This simplification happens after assuming a solution form like \( y(t) = \sum_{n=0}^\infty a_n(t-t_0)^{n+r} \) and substituting it into the differential equation. Solving this indicial equation provides the roots or exponents that describe the behavior of solutions near the singular points.

Exponents at Singular Points

Exponents at singular points are derived from the roots of the indicial equation. They play a critical role in defining the solution's behavior near each singularity. For the given equation, both regular singular points \( t = 2 \) and \( t = -2 \) produce the same indicial equation with roots \( r = 0 \) and \( r = 1 \).
These exponents inform us that at each singular point, the solution can be represented in terms of these roots:

• A typical solution is often a linear combination of terms like \((t-t_0)^r\) where \( r \) takes on these values.
• The presence of two distinct roots generally results in two linearly independent solutions.

Thus, knowing these exponents is essential as they guide the form and construction of the general solution in the vicinity of the singular points.

Characteristic Polynomial

In the context of differential equations, a characteristic polynomial is often associated with finding the indicial equation. It can refer to the polynomial derived in cases where we assume a series solution and collect like terms. This polynomial helps determine crucial parameters like the roots that define possible solution behaviors.
For our differential equation, the simplified indicial equation

• \[r^2 = 0\]

represents a characteristic polynomial for both regular singular points \( t = 2 \) and \( t = -2 \). This characteristic polynomial's simplicity, a perfect square, tells us there is repeated root behavior.
Understanding these characteristics assists in predicting singular behavior and constructing accurate series solutions for differential equations. Therefore, it's an indispensable part of analyzing differential equations, particularly around regular singular points.