Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter your answers as a comma-separated list. Include both real and complex singular points. If there are no singular points in a certain category, enter NONE.) x3y'' + 7x2y' + 4y = 0

x³y'' + 7x²y' + 4y = 0

has irregular singular point at x = 0.

Step-by-step 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.

Step-by-step explanation:

The given differential equation is:

[tex]x^3y'' + 2x^2y' + 4y[/tex]

the main task here is to determine the singular points of the given differential equation and Classify each singular point as regular or irregular.

So, for a regular singular point ; [tex]x=x_o[/tex] is located at the first power in the denominator of P(x) likewise at the Q(x) in the second power of the denominator. If that is not the case, then it is termed as an irregular singular point.

Let first convert it to standard form by dividing through with x³

[tex]y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0[/tex]

[tex]y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0[/tex]

The standard form of the differential equation is :

[tex]\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0[/tex]

Thus;

[tex]P(x) = \dfrac{2}{x}[/tex]

[tex]Q(x) = \dfrac{4}{x^3}[/tex]

The zeros of [tex]x,x^3[/tex] is 0

Therefore , the singular points of above given differential equation is 0

Classify each singular point as regular or irregular.

Let p(x) = xP(x) and q(x) = x²Q(x)

p(x) = xP(x)

p(x) = [tex]x*\dfrac{2}{x}[/tex]

p(x) = 2

q(x) = x²Q(x)

q(x) = [tex]x^2 * \dfrac{4}{x^3}[/tex]

q(x) =[tex]\dfrac{4}{x}[/tex]

The function (f) is analytic if at a given point a it is represented by power series in x-a either with a positive or infinite radius of convergence.

Thus ; from above; we can say that q(x) is not analytic at x = 0

[tex]Q(x) = \dfrac{4}{x^3}[/tex] do not satisfy the condition,at most to the second power in the denominator of Q(x).

Thus, the point x =0 is an irregular singular point