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Hello and welcome to problem 12 of chapter 2 section 4.
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Here are to state the region in the t .y plan or the hypotheses of theorem 2 .4 .2 are satisfied.
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Thus, there is a unique solution to reach given initial point in the region.
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A differential equation, dydt is equal to the cotangent of t times y all over 1 plus y.
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And to solve this, we have to consider the hypotheses of theorem 2 .4 .2, to that if this cotangent of t times y term, which we're going to call f, and its partial derivative with respect to why, partial left, partial y, are continuous in a region, then that region will have a guaranteed unique solution if we're given an initial point in that region.
00:44
So let's start off by talking about when this will be continuous, and we'll have two points of discontinuity.
00:56
It will be this top part, cotangent of t, which is.
01:00
We know is 1 over tangent of t for to make things a little bit simpler and this bottom part which cannot be equal to zero let's start off with the simpler one we have 1 plus y is equal to 0 from here we realize that y being negative 1 gives us a a non -continuous answer so that's the first one the second one is we have to find out when the cotangent of t is continuous.
01:38
And to make this a little bit simpler, we're going to write this as 1 over the tangent of t, which again can be written as cosine of t over sine of t.
01:58
From here, we're just finding out when sine of t is equal to zero...