Re: Ivp
This is a pretty standard "exact differential" problem. First, the "Assume t> 0" is to avoid t= 0 which, because of the "1/t", would cause trouble. Of course there would be another solution if we were to "assume t< 0".
(I just noticed that I have used "x, y" where you have "t, y". Assume my "x" is your "t".)
If F(x, y) is a differentiable function of x and y, then its differential is \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy. As long as those partial derivatives are smooth, we must have the "mixed derivatives" the same: \frac{\partial^2 F}{\partial x\partial y}= \frac{\partial^2 F}{\partial y\partial x} (where the order of "\partial x" and "tex]\partial y[/tex]" indicates the order of differentiation).
Now, we can write "u(x,y)dx+ v(x,y)dy" which looks like a differential- but it may not be! (We would say it is not "exact".) If this were really a differential, We would have to have \frac{\partial F}{\partial x}=u and \frac{\partial F}{\partial y}= v then \frac{\partial^2 F}{\partial x\partial y}= \frac{\partial u}{\partial y}= \frac{\partial v}{\partial x}= \frac{\partial^2 F}{\partial y\partial x}.
That is what is called the "mixed partials test" and they are showing after "Since".
Because that is true, we know there exist some function, F(x,y) such that \frac{\partial F}{\partial x}= \frac{1}{x}+ 2y^2x. Integrating (with respect to x, treating y as a constant) F= ln(x)+ y^2x^2+ \phi(y). (Since we are treating y as a constant, the "constant of integration" may depend on y. That is the "\phi(y)" above.)
Now, differentiate F= ln(x)+ y^2x^2+ \phi(y) with respect to y: \frac{\partial F}{\partial y}= 2y^2x+ \frac{d\phi}{dy}. (Notice that the derivative of \phi is an ordinary derivative, not a partial derivative because \phi is a function of the single variable y.)
But we know, from the equation, \frac{\partial F}{\partial y}= 2yx^2- cos(y) so we must have 2y^2x+ \frac{d\phi}{dy}= 2yx^2- cos(y). Notice that the only terms involving x, 2yx^2, cancel! That had to happen because of the "mixed derivatives" condition. That leaves just \frac{d\phi}{dy}= -cos(y) so that \phi(y)= -sin(y)+ C (C really is a constant here) so that
F(x,y)= ln(x)+ y^2x^2- sin(y)+ C.
Since the differential equation was "dF= 0", F must be a constant: F(x,y)= ln(x)+ y^2x^2- sin(y)+C= c or just F(x, y)= ln(x)+ y^2x^2- sin(y)= C' where I have combined the two constants: C'= c- C.
Dropping the "F", which was not part of the original equation, ln(x)+ y^2x^2- sin(y)= C.
(Replacing x with your t:
ln(t)+ y^2t^2- sin(y)= C.)
(I suspect the "y^2ts" was a typo. It should be just "y^2t".)