Differential Equations: Autonomous Systems, Stability, Trajectories

This assignment focuses on solving differential equations, analyzing the stability of autonomous systems, and sketching trajectories. It involves finding critical points for given systems of equations and determining their stability using Jacobian matrices and eigenvalues. The solution covers three questions, each requiring the identification of critical points, stability analysis (stable node, unstable saddle point, stable center, stable spiral), and trajectory sketching. The final question involves determining if a system has periodic solutions. The document concludes with relevant references and is available on Desklib, a platform offering study tools and solved assignments for students.

Running head: DIFFERENTIAL EQUATIONS 1
Differential Equations
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DIFFERENTIAL EQUATIONS 2
Section 1
Problem Statement
A set of equations is provided for which it is required to find the critical point of the
given autonomous systems and analyze their type of stability.
Solutions
Question 1
dx
dt =x−xy
dy
dt =x+2 y
Equate the two equation above and solve for x and y to get the various critical points
x−xy=0 … … … … … ..1
x +2 y =0 … … …… … .. 2
Factoring the first equation we have:
x (1− y )=0
From the above equation we have either x=0 or X is not equal two zero (1-y=0). For X=0, and
substituting to equation 2 above we have:
0+2 y=0
Which gives, y is 0 as well. The first critical point is therefore the point (0,0).
For X is not equal to zero, we have (1-y=0), giving y as 1. Substituting to equation 2 above we
get:
x−2 ( 1 ) =0
Solving for x will give x as -2. The second critical point is therefore (-2, 1).

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To determine the stability of the above system, we begin by determining the Jacobean,
the n we find the Eigen values of the Jacobian at the critical points (Carter, 2003).
The Jacobean matrix is given by the following:
j ( x , y ) =
[ du
dx
du
dy
dv
dx
dv
dy ]
This gives us the following matrix:
j ( x , y ) =(− y+1 −x
1 2 )
To get the Eigen values at each critical point, we subtract from the above Jacobean matrix
a 2X2 matrix as below and equate it to zero.
For critical point 1 (0,0) we have:
[ 1 0
1 2 ] − [ λ 0
0 λ ]=0
[ 1−λ 0
1 2−λ ]=0
Solving the above matrix, yields the following equation:
λ2−3 λ+2=0
Solving the equation above gives the following,
λ2−2 λ− λ+2=0
λ ( λ−2 )−1 ( λ−2 )=0
( λ−1 ) ( λ−2 ) =0
From the above we have λ=1and λ=2

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Since the two Eigen values are positive, it means that the point is an unstable node hence
all the trajectories near the neighborhood of the point will be directed outward and away from the
point.
Using critical point 2 (-2,1), we have
[ 0 2
1 2 ] − [ λ 0
0 λ ]=0
[ −λ 2
1 2−λ ]=0
Solving the above matrix, yields the following equation:
λ2−2 λ−2=0
Solving the equation above, we have
From the above we have λ=1+ √3and λ=1− √3. This indicates that the Eigen values
have opposite signs, one is positive while the other is negative. Therefore, this is an unstable
saddle point hence the trajectories in the direction of the negative Eigen values, the Eigen vector
will initially approach the point but will diverge as they approach the region that will be
dominated by the positive Eigen values.
Question 2
dx
dt =−x−2 y
dy
dt =2 x + y
Equate the two equation above and solve for x and y to get the various critical points
−x−2 y=0 … …… … ….. 1
2 x+ y =0 … … …… … .. 2

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Factoring the first equation we have:
−x=2 y hence x=2 y
Substituting to equation 2 above yields:
2 (−2 y )+ y=0
−4 y + y=0 hence−3 y =0∧thus y=0
The critical point of the system is therefore (0,0)
The Jacobean of the two equations obtained as in question 1 will be:
j ( x , y ) =[−1 −2
2 1 ]
To get the Eigen values at the critical point, will have:
[−1 −2
2 1 ]− [λ 0
0 λ ]=0
[−1−λ −2
2 1− λ ]=0
Solving the above matrix, yields the following equation:
λ2+3=0
λ2=−3
The Eigen values will therefore be λ=i √3 since the root of -1 is represent as “I”
indicating imaginary. Since the Eigen value is pure positive imaginary, then the point is a stable
center meaning the trajectories will seem to be coming from the origin outwards.
Question 3
dx
dt =−x2+ y2
dy
dt =1−x
Equate the two equation above and solve for x and y to get the various critical points

DIFFERENTIAL EQUATIONS 6
−x2+ y2=0 … … … … … .. 1
1−x=0 … … … … … ..2
From equation 1:
( y−x ) ( y + x ) =0
From equation 2:
1−x=0 hence x=1
Substituting to equation 1 above yields.
y−1=0 hence y =1
Therefore, critical point 1 is (1,1).
On the other hand, the second part yields
y +1=0 hence y=−1
Giving the second critical point as (1, -1)
The Jacobean of the two equations obtained as in question 1 will be:
j ( x , y ) =
[−2 x 2 y
−1 0 ]
To get the Eigen values at the critical point 1 (1, 1), will have:
[−2 2
−1 0 ]− [ λ 0
0 λ ]=0
[−2−λ 2
−1 −λ ]=0
Solving the above matrix, yields the following equation:
λ2+2 λ+ 2=0
When the above equation is solved, the Eigen values obtained are λ=−1+i and λ=−1−i
. In this case, the real parts of Eigen values are negative, hence the point is a stable spiral

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meaning that all trajectories in the neighborhood of the point spiral into the point an ever
decreasing radius.
Using the critical point 2 (1, -1), we have
[ −2 −2
−1 0 ] − [ λ 0
0 λ ]=0
[ −2−λ −2
−1 −λ ]=0
Solving the above matrix, yields the following equation:
λ2+2 λ−2=0
When the above equation is solved, the Eigen values obtained are λ= √3−1 and
λ=−1− √ 3. In this case, the Eigen values have opposite signs. That is positive and the other
negative, this means that the point can be described as an unstable saddle point, hence
trajectories in the direction of the negative Eigen values, the Eigen vector will initially approach
the point but will diverge as they approach the region that will be dominated by the positive
Eigen values (Mattuck, 2011).
Section 2
Problem Statement
This section requires as to sketch the trajectory of the given system and determine
whether the system has any periodic solutions.
Question 4
dx
dt = y
dy
dt =−12 x −6 x2
Solution

DIFFERENTIAL EQUATIONS 8
We find the resting point of the above set of equations by first equating them to zero. That is:
y=0 … … … … … … .1
−12 x−6 x2=0 … … … … … … .2
Getting the derivative of the above two equations yields,
y=0
−12−6 x=0
Solving for X in the second derivative yields:
−6 X =+12 hence X =−2
The resting point is therefore (-2,0)
The next step is to classify the resting point (-2,0) by writing the equation in terms of dy
dx . From
the above equations
dx
dt = y
While
dy
dt =−12 x −6 x2
To write the equation in terms of dy
dx we cross multiply the above equations as below:
dy
dx = dy
dt X dt
dx =−12 x−6 x2
y
The above equation when solved yields
∫ ydy =∫−12 x −6 x2 dx
Solving for both sides give,
y2
2 =−2 x3−6 x2 +C
y2=−4 x3−12 x2 +2 C

DIFFERENTIAL EQUATIONS 9
We replace C with 1 to get the trajectory equation as below
y2=−4 x3−12 x2 +2
y= √−4 x3−12 x2 +2
The plot of the above trajectory is as below:
The above system has no periodic solutions.

DIFFERENTIAL EQUATIONS 10
References
Carter, W. C. (2003). Stability of critical points. Retrieved from http://pruffle.mit.edu/3.016
2005/Lecture_25_web/node2.html
Mattuck, A. (2011). Ordinary differential equations. Retrieved from
https://math.mit.edu/~jorloff/suppnotes/suppnotes03/gs.pdf