Singular Solution

A function (x) is known as the singular solution of differential equation F (x, y, y’) = 0 if the uniqueness of the solution is disrupted at each point of the domain of the equation. Geometrically, this implies that more than one integral curve with the common tangent point passes through each point (x₀, y₀).

Note: Sometimes, the weaker singular solution definition is used when the uniqueness of the solution of the differential equation is disrupted only at some point. 

A singular solution of a differential equation is not defined by the general integral, that is it cannot be derived from the general solution of any specific value of Constant C.

Singular Solution Example Problem

Suppose the following equation is asked to be solved:

\[(y')^{2} - 4y = 0\]

It can be easily seen that the general solution of the differential equation is given by the function \[y = (x+c)^{2}\].

Graphically, the image is represented by the family of parabolas as shown below:

(Image will be Updated soon)

Besides this, the function y = 0 also satisfies the differential equation. However, this function is not included in the general solution. As more than one integral point passes through each point of the straight line y = 0, the uniqueness of the solution is violated at this point and hence, it is known as a singular solution of the differential equation.

Regular Singular Point

Let us consider the following differential equation

A(x)y’’ + B(x)y’ + C(x)y = 0

If both A and B are polynomials, them regular singular point x₀ is a singular point for which 

\[\lim_{x\rightarrow x_{0}}\] (x - x₀) \[\frac{B(x)}{A(x)}\] is infinite and \[\lim_{x\rightarrow x_{0}}\] \[(x - x_{0})^{2}\] \[\frac{C(x)}{A(x)}\] is finite.

For more general function other than polynomial, x₀ is a regular singular point if is considered to be a singular point with 

(x - x₀)\[\frac{B(x)}{A(x)}\] and \[(x - x_{0})^{2}\] \[\frac{C(x)}{A(x)}\] are analytic at x = x₀ 

A singular point that is not a regular singular point is known as an irregular singular point. 

Singular Matrix Solution

A matrix is said to be singular if its determinant is equal to 0. 

For S example, \[\begin{bmatrix}2 & 6 \\1 & 3 \end{bmatrix}\] is a singular matrix as 2.3 - 6.1 = 0

Note: A singular matrix is non-invertible which implies that its inverse does not exist. Let us understand what inverse does not exist for singular matrices.

The inverse of matrix X is given as:

X' = \[\frac{adjoint(X)}{|X|}\]

In case of singular matrix, |X| = 0

The denominator needs to be 0 in the case of a singular matrix, and that is not defined.

Therefore, the inverse of the singular matrix does not exist.

Solving Singular Matrix

Determine whether the given matrix is singular or not

\[\begin{bmatrix}2 & 4 & 6 \\2 & 0 & 2 \\6 & 8 & 14 \\\end{bmatrix}\]

Solution:

\[\begin{bmatrix}2 & 4 & 6 \\2 & 0 & 2 \\6 & 8 & 14 \\\end{bmatrix}\]

The determinant for the given matrix is calculated as:

2(8 2 - 14 0) - 4 (2 14 - 6 2) + 6(2 8 - 60) 

= 2(16 - 0) - 4 (48 - 12) + 6(16 - 0)

= 2(16) - 4(16) + 6(16)

= 32 - 64 + 96

= 0

As the determinant is 0, hence the given matrix is singular.

Singular Solution Example Problems With Solution

1. Find the singular point of the differential equation and classify them as regular or irregular?

(x² - 9)²y’’ + ( x + 3)y’ + 2y = 0

Solution:

Here, 

x = 3 is an irregular singular point 

x = -3 is a regular singular point.

2. Find the regular singular point of differential equation 

(1 - x²)y’’ - 2x y’ + (+ 1)y = 0

Where,

is a real constant

Solution:

As we know:

\[\frac{(x-1) Q(x)}{P(x)}\] = \[\frac{2x}{1+x}\] = \[\frac{(x-1)^{2} R(x)}{P(x)}\] = \[\frac{(x-1) \alpha(\alpha + 1)}{1+x}\]

Furthermore, the limits given below are infinite,

\[\lim_{x \to 1}\] \[\frac{(x-1) Q(x)}{P(x)}\]= 1 , \[\lim_{x \to 1}\] \[\frac{(x-1) R(x)}{P(x)}\]= 0

Therefore, we can conclude that x₀ = 1 is a regular singular point.