Question

Find critical points from dy/dx = yln(y+2). I know that we have to set: yln(y+2) = 0, but don't know what to do to get started after that....

Answer 1

What's the original question?

Answer 2

find the critical points and phase portrait
of the given autonomous first-order differential equation.
Classify each critical point as asymptotically stable, unstable,
or semi-stable.

Answer 3

I just can't get past the first part where it say s critical points....

Answer 4

the critical points would be the y values that make the equation y(ln(y+2)=0 or if the y values make the equation impossible

Answer 5

I can't figure that out with ln..

Answer 6

it should be y=0 and y=-2

Answer 7

ln(0) does not exist therefore it is a critical point

Answer 8

yln(y+2) = 0 when y = 0 or ln(y+2) = 0 or y = -1

Answer 9

why is y=-1 a critical point

Answer 10

oh nvm

Answer 11

to find if stable or unstable, take deriviative of the yln(y+2), evaluated at each critical point and check if > 0 or < 0. If < 0 then stable, if > 0 unstable

Answer 12

thanks.