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Diff Eq 5.2

 

<== Differential Equations
Number Question Answer
See the work
1. Solve the system
\displaystyle rac{dx}{dt} =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ -8 0 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-8 0
x

with the initial value
x(0) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
14
.

x(t) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
.
x(t) =


.
The work
2. Consider the system of differential equations
\begin{array}{rcl}  \displaystyle rac{dx}{dt} & = & -1.2 x + 1 y, \cr  			    &   & 		  \cr \displaystyle rac{dy}{dt} & = & 1.25 x  - 3.2 y.  nd{array}
For this system, the smaller eigenvalue is and the larger eigenvalue is .
The solution to the above differential equation with initial values x(0) = 2, \,\, y(0) = 3 is
x(t) = ,
y(t) = .
Smaller Eigenvalue= -3.7
Larger Eigenvalue= -0.7
x(t) =
y(t) =2.5*(2/3)*[e^(-3.7*t)]+(4/3)*[e^(-0.7*t)]
The work

Point Cost: 3
3. Consider the systems of differential equations
\begin{array}{rcl}  \displaystyle rac{dx}{dt} & = & 0.4 x + 0.5 y, \cr  			    &   & 		  \cr \displaystyle rac{dy}{dt} & = & 1.5 x  - 0.6 y.  nd{array}
For this system, the smaller eigenvalue is and the larger eigenvalue is .
The solution to the above differential equation with initial values x(0) = 3, \,\, y(0) = 4 is
x(t) = ,
y(t) = .

 


Smaller Eigenvalue= -1.1
Larger Eigenvalue= 0.9
x(t) =
y(t) =
The work

Point Cost: 3
4. Consider the systems of differential equations
\begin{array}{rcl}  \displaystyle rac{dx}{dt} & = & 0.3 x  - 0.4 y, \cr  			    &   & 		  \cr \displaystyle rac{dy}{dt} & = & -0.4 x + 0.9 y.  nd{array}
For this system, the smaller eigenvalue is and the larger eigenvalue is .
The solution to the above differential equation with initial values x(0) = 4, \,\, y(0) = 2 is
x(t) = ,
y(t) = .
Smaller Eigenvalue= 0.1
Larger Eigenvalue= 1.1
x(t) =
y(t) =
The work

Point Cost: 3
5. Solve the system
\displaystyle rac{dx}{dt} =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 2 2 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
-4 -2
x

with x(0) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 6 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
9
.

Give your solution in real form.
x_1 = ,
x_2 = .

x_1 =6*cos(2*t)+15*sin(2*t)
x_2 =-6*cos(2*t)-6*sin(2*t)+15*cos(2*t)-15*sin(2*t
The work
6. Solve the system
\displaystyle rac{dx}{dt} =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ -3 -4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
4 -3
x

with x(0) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 4 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
9
.

Give your solution in real form.
x_1 = ,
x_2 = .
x_1 =
x_2 =
The work

Point Cost: 3
7. Solve the system
\displaystyle rac{dx}{dt} =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ -6 -3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
15 6
x

with x(0) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 3 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
6
.

Give your solution in real form.
x_1 = ,
x_2 = .
x_1 =
x_2 =
The work

Point Cost: 3
8. Solve the system
\displaystyle rac{dx}{dt} =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 4 -5 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
5 4
x

with x(0) =
\setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight[ 9 \setlength{\arraycolsep}{0in}\left.\begin{array}{c}\! \\! \\! \nd{array}
ight]
7
.

Give your solution in real form.
x_1 = ,
x_2 = .
x_1 =
x_2 =
The work

Point Cost: 3