Drawing a phase portrait

Question

Can someone help me draw a phase portrait for $ y'' = -\log(y)-1 $? A full example will be much appreciated since when we talked about it in class, I really didn't understand it. Thank you very much!

Answer 1

Hint: First rewrite the system as a first order system by introducing $y' = z$ as an intermediate variable to obtain:

$$y'=z$$ $$z'=-\log y -1.$$

Now, divide both equations to obtain

$$\dfrac{dz}{dy}=\dfrac{-\log y -1}{z} \implies zdz = \left[-\log y -1\right]dy.$$

Integration will lead to:

$$\dfrac{1}{2}z^2=-\left[y\log y -y \right] -y+\dfrac{c}{2}$$ $$z^2=c-y\log y $$

Using the inital conditions $(z(t=t_0),y(t=t_0))=(z_0,y_0)$ $$z_0^2=c-y_0\log y_0 \implies c=z_0^2+y_0\log y_0.$$

So in total, we obtain:

$$z^2=z_0^2+y_0\log y_0 -y\log y $$ $$z^2=z_0^2-\log \left[\dfrac{y^y}{y_0^{y_0}} \right].$$

Now, it is up to you to look how the solutions look like as plots.