$yy'=\sin(t),y(0)=1$ phase portrait I need to draw a phase portrait for the equation
$y(t)y'(t)=\sin(t)$ with the initial condition $y(0)=1$. So far i've found that $y(t)= \sqrt{3-2\cos(t)}$ and $y'(t)=\frac{sin(t)}{\sqrt{3-2\cos(t)}}=\frac{sin(t)}{y(t)}$. I am not sure how to draw this phase portrait though, and i don't want to rely on computer tools too much, if someone could walk me through on how to draw the phase portrait for this that would really help. Please show steps.
 A: Normally, the horizontal axis is $y$ and the vertical axis is $y'$.
The two points where $y'(t)=0$ are $t=0$ and $t=\pi$, which is when $y=1,\sqrt{5}$ respectively.  So you have two points $A=(1,0)$ and $B=(\sqrt{5},0)$.  You are almost done.  There is a curve above the $y'$ axis connecting $A$ to $B$, and a curve below the $y'$ axis connecting $B$ to $A$.  The result looks like an approximate ellipse.
The direction of this path is clockwise because $y'>0$ in the upper half plane, so $y$ is increasing in the upper half plane, and $y'<0$ in the lower half plane, so $y$ is decreasing in the lower half plane.
A: $$
y = \sqrt{3-2\cos t}\implies \cos t = \frac{3-y^2}{2}
$$
we can see that 
$$
yy' = \sin t = \pm\sqrt{1-\cos^2 t} = \pm\sqrt{1- \left(\frac{3-y^2}{2}\right)^2}
$$
thus maybe you can use
$$
y' = \pm\frac{\sqrt{1- \left(\frac{3-y^2}{2}\right)^2}}{y}
$$
this crosses the $y'=0$ line at
$$
\left(\frac{3-y^2}{2}\right)^2 = 1 \\
\frac{3-y^2}{2} = \pm 1 \implies y^2 = 3\mp2 \implies y = \left\{\pm\sqrt{5},\pm 1\right\}
$$
since you can not cross $y=0$ unless you want a blow up solution then there is a symmetry about that axis.
