Show that all solutions remain in the interval for all time I really have no idea on how to get started with these, there's no similar example in my book. Do I need to compute $\frac{dy}{dx}$? Any help would be greatly appreciated. Maybe there's just some theorem I am not aware of
Show that all solutions $x(t),y(t)$ which start inside the given interval ($x>0,y>0$ for part a,$x>0$ for part b, $x^2+y^2=1$ for part c) must remain there for all time:
a. $$\frac{dx}{dt}=x^2+ysinx$$ $$\frac{dy}{dt}=-1+xy+cosy$$ $$x>0,y>0$$
1st attempt: $\frac{dy}{dx}=\frac{-1+xy+cosy}{x^2+ysinx}$
I can't remember how to solve for this since it's not separable
b. $$\frac{dx}{dt}=y(e^x-1)$$ $$\frac{dy}{dt}=x+e^y$$ $$x>0$$
1st attempt: $\frac{dy}{dx}=\frac{x+e^y}{y(e^x-1)}$
c. $$\frac{dx}{dt}=-1-y+x^2$$ $$\frac{dy}{dt}=x+xy$$ $$x^2+y^2=1$$
1st attempt: 
I just substituted...I don't think this is right though
$\frac{dx}{dt}=-1-y+x^2$
$\frac{d(x^2+y^2)}{dt}=1-y+(x^2+y^2)^2$
$\frac{dy}{dt}=x+xy$
$\frac{d(x^2+y^2)}{dt}=x+x(x^2+y^2)$
hint: compute $\frac{d(x^2+y^2)}{dt}$

Hints.
In a. If $x$ or $y$ vanishes for some $t_0$, then they stay zero, and due to uniqueness, they are zero everywhere - contraction.
In b. Note that if $x(t_0)= 0$, for some $t_0$, then $x\equiv 0$, again due to uniqueness.
In c. $(x^2+y^2)'=-2x(x^2+y^2-1)=0$.
