Question

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.) x^3y'' 2x^2y' 4y

Answer 1

Answer:

Step-by-step explanation:

The given differential equation is:

$x^3y'' + 2x^2y' + 4y$

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 ;  $x=x_o$ 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

$y'' + \dfrac{2x^2y'}{x^3} + \dfrac{4y}{x^3} =0$

$y'' + \dfrac{2y'}{x} + \dfrac{4y}{x^3} =0$

The standard form of the differential equation is :

$\dfrac{d^2y}{dy} + P(x) \dfrac{dy}{dx}+Q(x)y =0$

Thus;

$P(x) = \dfrac{2}{x}$

$Q(x) = \dfrac{4}{x^3}$

The zeros of $x,x^3$  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) = $x*\dfrac{2}{x}$

p(x) = 2

q(x) = x²Q(x)

q(x) = $x^2 * \dfrac{4}{x^3}$

q(x) =$\dfrac{4}{x}$

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

$Q(x) = \dfrac{4}{x^3}$  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