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differential equations with linear algebra

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1a) the Eigen values and the respective Eigen vectors of
2 2
2 1
÷
are 3, -2 and
2 1
,
1 2
So, the solution of the system of differential equations is
2
1 2
2 1
1 2
t t
x
c e c e
y
= +
Writing the differential equations, we see that
2 2
2
dx
x y
dt
dy
x y
dt
= +
=
Equating the right hand sides of the equations to zero, we get
0, 0x y= =
as the unique equilibrium
point.
The phase portrait is
It is easy to follow from the phase portrait that the equilibrium point (0, 0) is an unstable node.
1b) the Eigen values and the respective Eigen vectors of
1 1
1 1
÷
are
1 ,1i i+
and
,
1 1
i i
So, the solution of the system of differential equations is
( ) ( )
1 2
cos sin
1 1
t
x i i
e c t c t
y
= +
Writing the differential equations, we see that
dx
x y
dt
dy
x y
dt
= +
= − +
Equating the right hand sides of the equations to zero, we get
0, 0x y= =
as the unique equilibrium
point.
The respective phase portrait is

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From the phase portrait, we follow that the equilibrium point is an unstable spiral point.
1c) the Eigen values and the respective Eigen vectors of
0 9
9 0
÷
are
and
,
1 1
i i
So, the solution of the system of differential equations is
( ) ( )
1 2
cos 9 sin 9
1 1
x i i
c t c t
y
= +
Writing the differential equations, we see that
9
9
dx
y
dt
dy
x
dt
=
= −
Equating the right hand sides of the equations to zero, we get
0, 0x y= =
as the unique equilibrium
point.
the phase portrait is
From this figure, we follow that the equilibrium point (0, 0) is the stable spiral.
2) Consider the system of linear differential equations in the form of matrices
3 0 4
0 2 0
0 0 3
=
x x
The respective Eigen values and the Eigen vectors are 2, -3, 3 and
0 2 1
1 , 0 , 0
0 3 0
Since all the Eigen values and the Eigen vectors are real, the general solution is

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2 3 3
1 2 3
0 2 1
1 0 0
0 3 0
t t t
x
y c e c e c e
z
= + +
3) Consider the initial value problem
1 3
3 1
=
x x
and
[ ]
2
0
1
=
x
The Eigen values and the respective Eigen vectors are
1 1
4, 2 & ,
1 1
So, the general solution is
4 2
1 2
1 1
1 1
t t
x
c e c e
y
= +
Substitute the initial condition that
[ ]
2
0
1
=
x
in this, we get
( ) ( )
4 0 2 0
1 2
2 1 1
1 1 1
c e c e
= +
From this, we get
1 2
1 2
2
1
c c
c c
= −
+ =
Solving these linear equations, we get
1 2
1 3
,
2 2
c c
= =
Thus, the particular solution of the initial value problem is
4 2
1 1
1 3
1 1
2 2
t t
x
e e
y
= +
4) Consider the system of differential equations
2 1
9 4
=
x x
and the respective Eigen values and
eigen vectors are
1, 1
and the respective eigen vectors are
1 0
,
3 0
Since the algebraic multiplicity is more than the geometric multiplicity, we write the general solution
to be
( )
1 2
1
3
t
x
c c t e
y
= +
5a) consider the system of differential equations
2 2 2
0 1 1
4 8 3
=
x x
The Eigen values and the respective Eigen vectors are 2, 3, 1 and
9 2 2
4 , 1 , 1
4 2 0
Observe that the Eigen values are unequal and so, the respective Eigen vectors are linearly
independent.
The matrix with linearly independent vectors or Eigen vectors as the columns is
9 2 2
4 1 1
4 2 0
P
=

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Anonymous
Really great stuff, couldn't ask for more.

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