$yy'=\sin(t),y(0)=1$ phase portrait

Question

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.

Answer 1

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.

Answer 2

$$ y = \sqrt{3-2\cos t}\implies \cos t = \frac{3-y^2}{2} $$ 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.