Can you do problems 2.2.2, 2.2.3, 2.24 Only I did problem 2.2.1 as an example but can you please show the rest in a similar manner? You need to draw the phase portrait and solve for the fixed points and the analytical solution

2.2 Fixed Points and Stability Analyze the following equations graphically. In each case, sketch the vector eld on the real line, nd all the xed points, classify their stability, and sketch the graph of A'U') for different initial conditions. Then try for a few minutes to obtain the analytical solution for x0): if you get stuck, don't try for too long since in sev- eral cases it's impossible to solve the equation in closed form! 2.2.1 2; : 4x3 16 2.2.2 1 :19.\" 2.2.3 2?; = x x3 2.2.4 at : 6)\" sin 2: ( 2) ( page 36 ) 2. 2.1 , 2.2.2 ; 2.2. 3 ; 2.2.4 2. 2.1 x ( + ) = 4x 2-16 Phase - line portrait 4x 2- 16:0 4x 1 = 16 -z x 2 = 4 X = + 2 We obtain two ( 2 ) fixed points : *=2 and * =- 2 We then determine stability of the points by testing values aroun points to odetermine the sign of x " ( + ) at the point . We do this as follows : X' ( - 3 ) = 4( - 3) 2 - 16 20 - so we draw an increase mare Repeat with values [-2, 2] and [2, to ] to get: x' ( - 1 ) = 4 ( -1) 2 - 16 co , so we draw a durcase mark ." x' ( 3 ) = 4 ( 3 ) ' - 16 >o , so we draw an increase mark . As such, we are able to determine stability X =-2 is an attractor so this point is stable, * = 2 is a repeller so this point is unstable . The analytical solution for * ( + ) : dx 4 x 2 -16 . Rearrange , at dx = ( 4 x 2 - 16 ) dt dx = dt 4 x 2-16 Take the integral of both sides dx = t 4X 2-16 + + ( 7 2 ) - ( xtz ) dx =t dx x 1 - 4 =t 16 X - 2 ax - x +2 ax ) =t ax =t " ( In | x - 2 1 - Inl x + 2]) +c = t ( X-2) (x+2 )