# Phase line - differential equations..

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.

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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 $$y=\sqrt{3} \tan \left( C+\sqrt{3} x\right)-1$$