#$&*
course Mth 279
5/17
Query 26 Differential Equations*********************************************
Question: Find the solutions to y ' = A y when
A = [ 5, 3; -4, -3 ]
and
y(1) = [2; 0].
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Your solution:
p(r) = det[A-rI]
[ 5-r, 3; -4, -3-r]
= (5-r)(-3-r) - (3)(-4)
= (-15 + 3r - 5r + r^2) - (-12)
= r^2 - 2r - 15 + 12
= r^2 - 2r - 3
= (r + 1)(r - 3)
The roots are:
r1 = -1, and r2 = 3
for r1 = -1, we solve (A-r1*I)*x1 = (A+I)*x1 = 0
[6, 3; -4, -2]*[x1; x2] = [0; 0]
Using elemetary row operations
(2/3)R1 + R2 -> R2
(1/6)R1 -> R1
[1, (1/2); 0, 0]*[x1; x2] = [0; 0]
this represents an equation:
x1 + (1/2)x2 = 0
This is not the only solution but will satisfy the equation.
x1 = [ 1, -2]
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for r2 = 3, we solve (A-r2*I)*x2 = (A - 3I)*x2 = 0
[2, 3; -4, -6]*[x1; x2] = [0; 0]
Using elementary row operations:
2R1 + R2 -> R2
(1/2)R1 -> R1
[1, (3/2); 0, 0]*[x1; x2] = [0; 0]
examining this we get the equation:
x1 + (3/2)x2 = 0
Satisfying the equation:
x2 = [ -3; 2]
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solutions:
(r1, x1) = (-1, x1)
y1(t) = e^(-t) * [ 1; -2] = [ e^(-t); -2e^(-t)]
(r2, x2) = (3, x2)
y2(t) = e^(3t) * [ -3; 2] = [ -3e^(3t); 2e^(3t)]
y(t) = c1*y1(t) + c2*y2(t)
= [e^(-t), -3e^(3t); -2e^(-t), 2e^(3t)]*[ c1; c2]
W(t) at t =1
W(1) = [ 1/e, -3e^3; -2/e, 2e^3] = 2e^2 + 6e^2 = 8e^2
which is not zero so this is a fundamental set.
[ 1/e, -3e^3; -2/e, 2e^3]*[ c1; c2] = [ 2, 0]
psi(1)^(-1) = 1/(8e^2) * [ 2e^3, 3e^3; 2/e, 1/e] = [ e/4, (3/8)e; (1/4)e^(-3), (1/8)e^(-3)]
psi(1)^(-1) * y(1) = C
[ e/4, (3/8)e; (1/4)e^(-3), (1/8)e^(-3)] * [ 2; 0] = [ e/2; (1/2)e^(-3)]
y(t) = (e/2)y1 + (1/2)e^(-3) * y2
= [ (e/2)e^(-t), (1/2)e^(-3)*(-3e^(3t)); (e/2)(-2e^(-t)), (1/2)e^(-3)*(2e^(3t))]
= [ (1/2)e^(1-t) - (3/2)e^(-3)*(-3e^(3t)); -2e^(1-t) + e^(3t - 3)]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
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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:
P(r) = det[A-rI] = [ 4-r, 2, 0; 0, 1-r, 3; 0, 0, -2-r]
=(4-r)[ 1-r, 3; 0, -2-r] - 2[ 0, 3; 0, -2-r] + 0[ 0, 1-r; 0, 0]
= (4 - r)(1 - r)(-2 - r)
= (-r + 4)(-r + 1)(-r - 2)
= (-r + 4)(r - 1)(r + 2)
= - (r - 4)(r - 1)(r + 2)
r1 = 4
r2 = 1
r3 = -2
For r1 = 4, we solve (A-r1*I)*x1 = (A-4I)*x1 = 0
[0, 2, 0; 0, -3, 3; 0, 0, -6]*[x1; x2; x3] = [0; 0; 0]
using elemetary row operations
(3/2)R1 + R2 -> R2
(1/2)R1 -> R1
2R2 + R3 -> R3
(1/3)R2 -> R2
[0, 1, 0; 0, 0, 1; 0, 0, 0]*[x1; x2; x3] = [0; 0; 0]
Since we have 2 equations:
x2 = 0
x3 = 0
Solution could be x1 = [0; 0; 0]
@&
x1 being unrestricted we could let x1 = 1.
The vector [1;0;0] is an eigenvector for this matrix, yielding [4; 0; 0]. The eigenvalue is 4.
*@
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for r2 = 1, we solve (A-r2*I)*x2 = (A-I)*x2 = 0
[ 3, 2, 0; 0, 0, 3; 0, 0, -3]*[x1; x2; x3] = [0; 0; 0]
Using elementary Row operations:
R2 + R3 -> R3
(1/3)R2 -> R2
(1/3)R1 -> R1
[1, 2/3, 0; 0, 0, 1; 0, 0, 0]*[x1; x2; x3] = [0; 0; 0]
This yeilds equations:
x1 + (2/3)x2 = 0
x3 = 0
This solution will satisfy the equations:
x2 = [ 2; -3; 0]
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for r3 = -2, we solve (A-r3*I)*x3 = (A +2I)*x3 = 0
[ 6, 2, 0; 0, 3, 3; 0, 0, 0]*[ x1; x2; x3] = [0; 0; 0]
Using elementary row operations:
(-2/3)R2 + R1 -> R1
(1/6)R1 -> R1
(1/3)R2 -> R2
[1, 0, -1/3; o, 1, 1; 0, 0 , 0]*[x1; x2; x3] = [0; 0; 0]
This yeilds equations:
x1 + (1/3)x3 = 0
x2 + x3 = 0
A solution that satisfies both:
x3 = [ 1; -3; 3]
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solutions:
(r1, x1) = (4, x1)
y1(t) = e^(4t)*[ 0; 0; 0] = [ 0; 0; 0]
@&
You could have used any value for x1 other than zero, so there is no reason to have chosen 0.
The vector [1; 0; 0] yields solution
y1(t) = e^(4 t) [1; 0; 0]
and will result in a fundamental set.
*@
(r2, x2) = (1, x2)
y2(t) = e^t * [ 2; -3; 0] = [ 2e^t; -3e^t; 0]
(r3, x3) = (-2, x3)
y3(t) = e^(-2t) * [ 1; -3; 3] = [ e^(-2t); -3e^(-2t); 3e^(-2t)]
y(t) = c1y1(t) + c2y2(t) + c3y3(t)
= [ 0, 2e^t, e^(-2t); 0, -3e^t, -3e^(-2t); 0, 0, 3e^(-2t)]*[c1; c2; c3]
W(t) t = 0
W(0) = [ 0, 2, 1; 0, -3, -3; 0, 0, 3] = 0
since the wronskian = 0, this is not a fundamental set.
But, here is the rest of the solution:
[ 0, 2, 1; 0, -3, -3; 0, 0, 3]*[c1; c2; c3] = [ -1; 0, 3]
by inspection there are 3 easily solvable equations:
3c3 = 3
c3 = 1
-3c2 - 3c3 = 0
-3c2 - 3(1) = 0
c2 = -1
2c2 + c3 = -1
2(-1) + 1 = -1 check.
therefore:
[ c1; c2; c3] = [ 0; -1; 1]
y(t) = [ 0, 2e^t, e^(-2t); 0, -3e^t, -3e^(-2t); 0, 0, 3e^(-2t)]*[c1; c2; c3]
= [ 0, 2e^t, e^(-2t); 0, -3e^t, -3e^(-2t); 0, 0, 3e^(-2t)]*[ 0; -1; 1]
= [ 0, -2e^t, e^(-2t); 0, 3e^t, -3e^(-2t); 0, 0, 3e^(-2t)]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
3
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
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Self-critique rating:
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Question: Three tanks, each full of a solution and all of equal volume V, are each connected to each of the others by two pipes. Each tank also has a third pipe through which pure water flows into it, and a fourth through which water exits.
The flow rate r through every pipe is the same.
Write the system of equations that relates the quantities Q_1, Q_2 and Q_3 representing the amount of solute in each tank as a function of time.
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Your solution:
@&
The concentrations of solution in the tanks are, respectively, Q1 / V, Q2 / V and Q3 / V.
Consider tank 1. It receives pure water at rate r, while solution with concentration Q1 / V exits the system from this tank at rate r. It also loses solution at the same concentration to each of the other two tanks. It gains solution from the other two tanks, each at rate r, the solution from tank two having concentration Q2 / V and the solution from tank three having concentration Q3 / V. Thus the net rate at which the amount Q1 of solute in the first tank is changing is
(Q_1) ' = -3 Q_1 / V * r + Q_2 / V * r + Q_3 / V * r.
Similar analysis can be made for the concentrations Q_2 and Q_3.
*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:"
&#This looks good. See my notes. Let me know if you have any questions. &#