Trying to solve for $\frac{\mathrm{dy} }{\mathrm{d} t}=\mu-y^{2}$
where $\mu<0$
reframing the right hand side I get $-\left ( \left | \mu \right |+y^{2}\right )$
and solving via separating variables we have $\left ( \frac{-1}{\mu^{\frac{1}{2}}} \right )tan^{-1}\left ( \frac{y}{\sqrt{\mu}} \right )=t+c$
Things is my signs are wrongs. But I'm not sure where did I fugged up. Can someone help please?
As you did, setting $\mu=-|\mu|$ makes the differential equation to be $$\frac{dy}{dt}=-(|\mu|+y^2)$$ $$\frac{dy}{|\mu]+y^2}=-dt$$ Using the same change of variable $y=z \sqrt{|\mu|}$, $dy= \sqrt{|\mu|}\,dz$ leads to $$\frac{dz}{1+z^2}=-|\mu|\,dt$$ $$\tan^{-1}(z)=-|\mu|t+c$$ $$z=\tan(-|\mu|t+c)=-\tan(|\mu|t+c)$$ $$y=- \sqrt{|\mu|}\tan(|\mu|t+c)$$