Find a function $ f $ such that $ f(3) = 2 $ and
$ (t^2 + 1)f'(t) + [f(t)]^2 + 1 = 0 t \neq 1 $
for this problem, we are asked to find a function apps such that f of three equals two and t squared plus F. Prime of T plus f of t squared plus one equals zero, where T does not equal one. So what we can do here is we want to set this up as a separable equation. So let's see here, we first and write this as f prime of T plus Let's see one moment. What we can do first is isolate the F prime of T onto one side. So we'd have F prime of T equals negative F squared minus one, divided by t squared plus one. Now to isolate our variables, we can divide both sides by negative x squared minus one. So we get F prime over negative x squared minus one is equal to one over T squared plus one. Now integrating the right hand side, The integral of that T squared plus one. Uh that is going to come out to one moment here, that's going to come out to arc tanne of T plus a constant. And then integrating the left hand side. There, we are going to get, I don't know, let me double check this one second. Yeah, if we integrate the left hand side, we just end up getting a negative arc tanne of F. So to isolate f We can multiply both sides by negative one and then take the tan of both sides. So we get that F is going to equal tan of negative arc tanne of T plus C. So the last step here is to solve for initial condition. So we want when T equals three. For this to equal to We need 10 of Negative Arc Tanne of three plus C. To equal to which then means that we can take the ark tan of both sides again. So we get that negative arc tanne of three plus C Needs to Equal Arc Tanne of two, which in turn means that we need C. Two equal arc tanne of two Plus Arc Tanne of three, and that's going to be the most precise way to communicate that.