#$&*
course Mth 279
8/5
Query 26 Differential Equations*********************************************
Question: Find the solutions to y ' = A y when
A = [ 5, 3; -4, -3 ]
and
y(1) = [2; 0].
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Your solution:
lamda= l
det(A - lamdaI) = 0
det([5,3;-4,-3] - lamda[1,0;0,1])
det[5-l, 3; -4, -3-l] = (5-l)(-3-l)-(3*-4)
l^2 -2l -3
(l-3)(l+1)
l=3 l=-1
l_1 = 3
plug in l into det()
[2,3;-4,-6]
[2,3;-4,-6][k_1; k_2] = 0
2k_1 + 3k_2 = 0
k_1 = [-3,2]
l_2 = -1
[6,3;-4,-2][k1;k2] = 0
6k_1 + 3k_2 = 0
3k_2= -6K_1
k_2 = -2k_1
let k_1 = 1
k_2 = [1;-2]
gen soln: y(t)=C1e^(l_1t)K_1 + C2e^(l_2t)K_2
y(t) = C_1[-3;2]e^(3t) + C_2[1;-2]e^-t
C_1[-3e^(3*1); 2e^(3*1)] + C_2[e^-1; -2e^-1] = [2;0]
-3e^3C_1 + e^-2C_2 = 2
2e^3C_1 + (-2e^-1)C_2 = 0
2e^3C_1 = 4
C_1 = 4/(2e^3) = 2e^-3
2e^3(2e^-3) - 2e^-1C2 = 0
4-2e^-2C_2 = 0
C_2 = 2e
y(t) = 2e^(-3)*[-3;2]e^(3t) + 2e*[1;-2]e^-t
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Find the solutions to y ' = A y when
A = [ 4,2,0; 0,1,3; 0,0, -2 ]
and
y(0) = [-1;0;3].
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Your solution:
det(A-lI) = 0
det( 4-l, 2, 0; 0, 1-l, 3; 0, 0, -2-l) = (4-l)((1-l)(-2-l)-(3*0)) - 2((0*(-2-l))-3*0)) + 0((0*0)-(1-l)(0))
0=(4-l)(1-l)(-2-l)
l_1=4 l_2=1 l_3=-2
l_1=4
[4-l, 2, 0; 0, 1-l, 3; 0, 0, -2-l][k1;k2;k3]=0
[0,2,0; 0, -3,3; 0,0,6][k1; k2; k3]=0
Not sure how to get this into rref without doing it on a calculator
@&
If you multiply your matrix equation out you get
2 k2 = 0
-3 k2 + 3 k3 = 0
6 k3 = 0
The first equation dictates that k2 = 0.
Substituting this into the second you get an equation that dictates k3 = 0.
There is no restriction on k1. So you're free to choose any value other than zero. The simplest value you can choose is 1.
So for lambda = 4, your solution vector is [1;0;0].
*@
l_2 = 1
[3,2,0; 0,0,3; 0,0,-3][k1;k2;k3] = 0
3k1+ 2k_2 = 0
3k_3 = 0
k_3 = 0
k_1 = -2/3 k_2
let k_2 = 3
k_2 = [-2;3;0]
l_3 = -2
[6,2,0;0,3,3; 0,0,0][k1;k2;k3]=0
6k_1 + 2k_2 = 0 -> k_1 = -1/3k_2 let k_2=3
3k_2 + 3k_3 = 0 -> k_3= -k_2
k_3 = [-1;3;-3]
gen soln:
y(t) = Ce^(l_1t)k_1 + C_2e^(l_2t)k_2 + C_3e^(l_3t)k_3
my only problem with this question is how to k_1???
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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Question: Three tanks, each full of a solution and all of equal volume V, are each connected to each of the others by two pipes. Each tank also has a third pipe through which pure water flows into it, and a fourth through which water exits.
The flow rate r through every pipe is the same.
Write the system of equations that relates the quantities Q_1, Q_2 and Q_3 representing the amount of solute in each tank as a function of time.
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Your solution:
Q’_1(t) = (rQ_3/V + rQ_2/V) - (rQ_1/V + rQ_1/V + rQ_1/V)
@&
Tank 1 has three pipes leading out of it, so your equation for Q_1 ' would include the term -3 Q_1 r / V:
Q_1' = -3(r/V)Q_1 + (r/V)Q_2 + (r/V)Q_3
Similar modifications would correct the two remaining equations.
*@
Q’_2(t) = (rQ_1/V + rQ_3/V) - (rQ_2/V + rQ_2/V + rQ_2/V)
Q’_3(t) = (rQ_1/V + rQ_2/V) - (rQ_3/V + rQ_3/V + rQ_3/V)
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
Self-critique (if necessary):
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Self-critique rating:
Self-critique (if necessary):
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#*&!
&#Good work. See my notes and let me know if you have questions. &#