00:01
In this problem we have a second order linear ordinary differential equation like this 1 minus x squared y double prime plus 3x y prime plus 8y equal to 0 and we are asked to determine whether each one of the singular points is a regular one or an irregular one.
00:37
So first we are going to determine the singular points and then a single point of the solution and then we are going to determine or decide on whether they are regular or irregular.
00:53
So the singular points are determined as the roots of the function, that is the coefficient of the y double prime turn so this is the function we are going to consider singular points let us denote the singular points in general by x sub 0 so we have 1 minus x0 squared equal to 0 so we see that the singular points are plus and minus 1.
01:36
We need to test them.
01:39
So let me write down the differential equation in the canonical form.
01:44
So i will have some function p of x times y double prime plus q of x times y prime plus r of x times y equal to zero.
01:58
And we should check if the following limits are both finite.
02:07
Let me first write down the limits.
02:12
We will evaluate this x minus x0 times q x divided by px as x goes to one of the single points on one at a time.
02:29
We need to see if this is finite and if the limit as x goes to x0, as x goes to x0, x minus x0 squared times r of x divided by p of x are are both finite so by by looking at the original the initial form of the differential equation we see that p of x is 1 minus x squared q of x is let's see 3x and r of x is just constant and given by 8.
03:21
So let us consider the single points 1 by 1.
03:25
We have x0 equal to 1.
03:29
So this is the first limit.
03:32
As x goes to 1, we have x minus 1 times 3x divided by 1 over x squared or limit.
03:45
As x goes to 1 we have x minus 1 times 3x divide by 1 minus x 1 plus x we have some cancellation here this guy and this guy and we will have minus 1 as the as an overall factor now the the term the term under the limit here is a nice one a regular one so we can simply insert the value x equal to 1 here and we will obtain...