Question

Find all singular points of the given equation and determine whether each one is regular or irregular. \((x+3) y^{\prime \prime}-2 x y^{\prime}+\left(1-x^{2}\right) y=0\)

Short answer

Answer: The singular point of the given differential equation is x = -3, and it is a regular singular point.

Step-by-step solution

Identify the leading coefficient of the highest derivative

The leading coefficient of the highest derivative, \(y''\), is \((x+3)\).

Find singular points by setting the leading coefficient to zero

To find the singular points, we set the leading coefficient to zero: \((x+3)=0\). Solving for x, we get: \(x = -3\). So, there is only one singular point at \(x=-3\).

Determine if the singular point is regular or irregular

A singular point is regular if all the coefficients of the equation have converging power series expansions around that point. Here, all coefficients are polynomials and they all converge everywhere (including at the singular point \(x=-3\)). Hence, the singular point \(x=-3\) is a regular singular point.

Key concepts

Singular Points

In differential equations, singular points are specific values of the independent variable where the equation might not behave normally. This happens when the coefficient of the highest derivative (in our case, the second derivative, or \(y''\)) becomes zero.
For the equation

• \((x+3)y'' - 2xy' + (1-x^2)y = 0\)

we see that the leading coefficient is \((x+3)\). To find singular points, we set this coefficient to zero, which gives us \(x = -3\).
Thus, \(x = -3\) is a singular point. Whenever solving similar equations, identifying these points is crucial as they often require special techniques to find solutions.

Regular Singular Point

A singular point is labeled as regular if the behavior of the differential equation around that point remains well-defined and manageable. Mathematically, this means the coefficients involved should have power series expansions that converge at the singular point.
In our exercise, the singular point occurs at \(x = -3\). The step involves examining each coefficient to see if a power series expansion around \(x = -3\) is possible. For

• \((x+3)y'' - 2xy' + (1-x^2)y = 0\)

all coefficients are polynomials, which naturally converge everywhere, including at \(x = -3\).
Therefore, \(x = -3\) qualifies as a regular singular point. Recognizing regular singular points is important as they often allow the use of series solutions, like Frobenius method, to solve the differential equation.

Irregular Singular Point

Irregular singular points behave differently and more erratically than regular singular points. At an irregular singular point, the differential equation's coefficients do not have a simple, converging power series expansion.
This irregular behavior makes finding solutions more complex and sometimes requires more advanced techniques beyond standard series solutions.
In contrast to our current exercise, where we confirmed \(x = -3\) is regular, an irregular point would involve coefficients that lead to divergence or undefined expansions at that specific point.
Understanding the distinction between regular and irregular points helps in choosing the right approach to handle the unique challenges posed by irregular singular points in solving differential equations.