For the differential equation: $$ y' = y(y+2) + 4 $$ I expanded it to: $$ y' = y^2 + 2y + 4 $$
I am asked to draw a phase line, and draw some typical solutions on the plane. However, there is no equilibrium solution to the equation above.
Is it correct that the phase line should have a upward pointing arrow(increasing function) for all values of y?.
Another thing is, I separated the variables but not sure if it is correct.
I get
$$ y^2 + 2y + 4 = ke^t $$ but am unsure on how to continue..
Thank you very much.
You face a separable differential equation. So, express $$x'=\frac{1}{y^2+2 y+4}$$ and integrate. This gives $$x=\frac{\tan ^{-1}\left(\frac{y+1}{\sqrt{3}}\right)}{\sqrt{3}}+C$$ and then $$y=\sqrt{3} \tan \left( C+\sqrt{3} x\right)-1$$