Answer:
x³y'' + 7x²y' + 4y = 0
has irregular singular point at x = 0.
Explanation:
Consider the differential equation
y'' + P(x)y' + Q(x)y = 0 ..................(1)
A point x = x_0 is called an ORDINARY POINT of (1) if both functions P(x) and Q(x) are analytic (differentiable) at x = x_0.
If the point x = x_0 is not an ordinary point, then it is a SINGULAR POINT.
There are two types of singular points:
REGULAR SINGULAR POINTS
IRREGULAR SINGULAR POINTS
A singular point x = x_0 of (1) is called REGULAR if both (x - x_0)P(x) and (x - x_0)²Q(x) are analytic at x = x_0.
otherwise, the singular point is called IRREGULAR.
EXAMPLE:
Determine if x = 0 is an ordinary point or singular point for the differential equation
x³y'' + 7x²y' + 4y = 0 .....................(2)
First, we rewrite (2) to be in form of (1) by dividing through by x³
y'' + (7/x)y' + (4/x³)y = 0
Comparing with (1)
P(x) = 7/x
Q(x) = 4/x³
At x = 0
P(x) = 7/0 = infinity
Q(x) = 4/0 = infinity
Both P(x) and Q(x) are nonanalytic, so the point x = 0 is not an ordinary point.
Again
(x - 0)P(x) = 7
(x - 0)Q(x) = 4/x
At x = 0
(x - 0)P(x) = 7
(x - 0)Q(x) = infinity
Only (x - 0)P(x) is analytic, so x = 0 is an IRREGULAR SINGULAR POINT.
To determine the singular point of a differential equation, all we need to do is find the point that satisfies the properties of a singular point as explained above.
x³y'' + 7x²y' + 4y = 0
has irregular singular point at x = 0.