I have the ODE $$dy - (y^{2}\sin(x))dx = 0$$ I have found the general solution of this (by transforming it into an exact equation and solving) to be $$y=\frac{1}{\cos(x)+c}$$ if $\cos(x)\neq c$.
If I now have the initial conditions $$y(0)=1$$ and $$y(\pi/2)=a$$ where $a=\pm1/2,\pm2/3,\pm1$, how do I plot these particular solutions in MATLAB?
My main reason for asking is that I get quite a different plot to my lecturer for the case of $y(0)=1$. For reference, the plots are over $0\leq x\leq4\pi$.
I get 
Whereas my lecturer gets 