00:01
This notation of y prime of t would be the same thing as saying d y d t, which i like this notation.
00:08
So when we're looking at t squared over y squared plus one, it's kind of like cross multiplying that i can multiply the y square plus one to the left side and cross multiplied the d t to the right side.
00:25
So we set that up that way so we can do the interval on both sides.
00:29
Then the integral on the left side, you're just adding one to the exponent and then multiplying by that.
00:33
The reciprocal in the new exponent, same thing over here, plus some constant.
00:41
So this would be your answer to a no matter what.
00:48
So then in matter b, there are several initial conditions, like one of them is negative one, one.
00:56
So when you plug in negative one for t, well, negative one cube stays negative times negative one third, and then one cubed still one times one -third plus one.
01:08
So what i can do is add one -third to the left side.
01:15
So that gives me a c value of, let's see, that's two -thirds plus three -thirds would be five -thirds.
01:25
So i could go back to this problem up here and rewrite it as one -third y -cued plus y is equal to negative, sorry, one -third t -cued, five thirds.
01:46
I'm sure there's other ways of writing it, but that's where i'm going to stop...