# $yy'=\sin(t),y(0)=1$ 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.

• i thought you could only draw phase portraits of autonomous system of differential equations: the differential equations that don't depend on $t$ explicitly which yours does. trouble is the direction field is time dependent and this causes trouble like trajectories can go intersect in space variables at different times without violating the uniqueness of solutions. – abel Dec 9 '14 at 14:10
• @abel You always can consider an extended phase space, substituting $t$ with $\theta$ and adding an equation $\dot{\theta}=1$. – Evgeny Dec 9 '14 at 15:06

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