Problem 1: Give the general solution to the DE system x l
=
− 2 1
− 1 − 4
) x and also give its phase-plane picture (i.e its direction field graph together with a few typical solution curves).
Solution: Characteristic equation
λ
2
+ 6
λ
+ 9 = (
λ
+ 3 ) = 0 → repeated root
λ
= − 3.
The single eigenvector v and a generalized eigenvector w such that the scalar component functions x
1
( t ) , x
2
( t ) of the general solution x
( A
=
−
λ
) x x
1
2
I ) w = v , and of the form x ( t ) = c
1 v e λ
2
( v t + of the given system x l
Eigenvector: t
+ v c
=
-
−
1
1
=
) w ) e
Ax
λ t are as follows:
1 )
Generalized eigenvector: w =
0
Thus, x
1
( t ) = ( c
1
+ c
2
+ c
2 t ) e
− 3 t and x
2
( t ) = ( − c
1
− c
2 t ) e
− 3 t
.
Problem 2: For each of the following linear systems, carry out the graphing program laid out in this session, that is:
(i) find the eigenvalues of the associated matrix and from this determine the geometric type of the critical point at the origin, and its stability;

Part I Problems and Solutions OCW 18.03SC
(ii) if the eigenvalues are real, find the associated eigenvectors and sketch the corre­ sponding trajectories, showing the direction of motion for increasing t ; then draw some nearby trajectories;
(iii) if the eigenvalues are complex, obtain the direction of motion and the approximate shape of the spiral by sketching in a few vectors from the vector field defined by the system. a) x l
= 2 x − 3 y , y l
= x − 2 y b) x l
= 2 x , y l
= 3 x + y c) x l
= − 2 x − 2 y , y l
= − x − 3 y d) x l
= x − 2 y , y l
= x + y e) x l
= x + y , y l
= − 2 x − y
Solution: Let x ( t ) =
x ( t ) ) y ( t ) throughout, and M be such that x l
( t ) = Mx ( t ) . Let M have eigenvalues
λ 1
,
λ 2
, with corr esponding eigenvectors x
1
,
2
. The general solution is thus
( t ) = c
1 1 e λ
1 t
+ c
2 x
2 e λ 2 t a) M =
2
1
− 3 )
− 2
, with eigenvalues ± 1 and eigenvectors
3
1
) and
1 )
1
. The system has a critical point at ( 0, 0 ) which is a saddle point.
For c
1
= 0 and as t → ∞
, x ( t ) = c
2 e
− t
1
1
)
→
0 )
0
Similarly, for c
2
= 0 and t → − ∞
, x ( t ) →
) 0
0
.
Thus the behavior near the saddle point looks like
2

Part I Problems and Solutions OCW 18.03SC b) M =
2 0 )
3 1
1 ) 0
, with eigenvalues 2, 1 and eigenvectors
3
,
1
)
. The system has an un­ stable node at ( 0, 0 ) .
As t → − ∞ all trajectories go to x 0.
Thus the behavior near the node looks like
For c
1 t → −
1 )
3 e
∞
, c
2
) 0
1 e t is dominant term, so the solutions are near the
2 t dominates, so solutions are parallel to
1
3
)
.
y -axis. For t → ∞
, c) M =
− 2 − 2 )
− 1 − 3
, with eigenvalues − 4, − 1 and eigenvecto rs
) ) 1
1
,
−
2
1
. The system has an asymptotically unstable node at (0,0). As t → ∞
, all trajectories go to x 0. The behavior near the origin looks like:
3

Part I Problems and Solutions OCW 18.03SC
For t → − ∞ ,
) 1
1 e
− 4 t dominates, so solutions are parallel to
) 1
1
; for t → ∞
,
-
−
2 )
1 e
− t dominates, so solutions come in to the origin asymptotic to the line with direction vector
2 )
− 1
. d) M =
(0,0).
1 − 2
1 1
)
, eigenvalues 1 ± i
√
2. The system then has an unstable spiral around
Near y = 0, x l ≈ x , so x is increasing where the spiral cuts the positive increases, so does e t
, so the spiral is outwards from the origin. x -axis. As y e) M =
1 1
− 2 − 1
)
The curves are ellipses, since dy dx
=
− 2 x − y x + y which integrates easily after cross-multiplying
4

Part I Problems and Solutions OCW 18.03SC to 2 x 2 + 2 xy + y 2 = c .
Direction of motion: For instance, at ( 1, 0 ) the vector field is x y l
= − 2, so motion is clockwise. l
= 1,
5

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