#$&*
course Mth 279
4/30 3
Query 26 Differential Equations*********************************************
Question: Find the solutions to y ' = A y when
A = [ 5, 3; -4, -3 ]
and
y(1) = [2; 0].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(A - `lambda*I)x = 0, going to abbreviate w/ lambda = `la
= [ 5 -`la, 3; -4, -3-`la ]
det[ 5 -`la, 3; -4, -3-`la ] = (5-`la)*(-3-`la) + 12 = [`la^2 -2`la - 15] + 12
= `la^2 -2`la - 3
eigenvalues =`la = -1,3
Eigenvector for `la = -1
[(5-(-1)), 3;-4, (-3-(-1))] *[y_1;y_2] = 0
6y_1 + 3y_2 =0
-4y_1 - 2y_2 = 0
eigenvector = r[1; -2], so we have y_1 =[1; -2]*e^(`la*t) = [1; -2]*e^(- t)
Eigenvector for `la = 3
[(5-(3)), 3;-4, (-3-(3))] *[y_1;y_2] = 0
@& Your solution is good throughout.
To avoid confusion I would use [x_1, x_2] to indicate the eigenvector rather that [y_1, y_2], in order to distinguish the coordinate of the eigenvector from the two solutions y_1 and y_2 of the equation. y_1 would be the usual notation for your vector solution [1, -2] e^-t and y_2 for the vector solution [1; -2/3] e^(3 t).*@
2y_1 + 3y_2 =0
-4y_1 - 6y_2 = 0
eigenvector = r[1; -2/3], so we have y_1 =[1; -2/3]*e^(`la*t) = [1; -2/3] *e^(3 t)
General solution is then
y(t) = c1[1; -2]*e^(- t) +c_2 [1; -2/3] *e^(3 t)
For y(1) = [2;0], we get
c1[1; -2]*e^(- 1) + [1; -2/3] *e^(3) = [2;0], solving
c_2 =(3/2*e^(3))
c_1 = (e/2)
Solution for init cond y(1) = [2;0] is then
y(t) = (e/2) [1; -2]*e^(- t) + (3/2*e^(3)) [1; -2/3] *e^(3 t)
= (1/2)[1;-2] + (3/2)[1;-2/3]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
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].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(A - `lambda*I)x = 0, going to abbreviate w/ lambda = `la
= [ 4 -`la, 2,0; 0,1-`la,3; 0,0,-2-`la ], which is upper triangular matrix making determinant easy to find
det[ 4 -`la, 2,0; 0,1-`la,3; 0,0,-2-`la ]= (4-`la)*(1-`la)*(-2 -`la)
eigenvalues =`la = 4,1,-2
Eigenvector for `la = 4
[ 4 - 4, 2,0; 0,1-4,3; 0,0,-2-4 ]*[y_1;y_2;y_3] = 0
2y_2 =0
-3y_2 + 3y_3 = 0
-6y_3 = 0
eigenvector = r[0;0;0], so we have y_1 =[0;0;0]*e^(`la*t) = [0;0;0]*e^(4 t)
Eigenvector for `la = 1
[ 4-1, 2,0; 0,1-1,3; 0,0,-2-1 ]*[y_1;y_2;y_3] = 0
3y_1 + 2y_2 =0
3y_3 = 0
-3y_3=0
eigenvector = r[1; -3/2;0], so we have y_2 =[1; -3/2;0]*e^(`la*t) = [1; -3/2;0]*e^( t)
Eigenvector for `la =-2
[ 4-(-2), 2,0; 0,1-(-2),3; 0,0,-2-(-2) ]*[y_1;y_2;y_3] = 0
6y_1 + 3y_2 =0
@& this would be 6 y1 + 2 y2 = 0
(see previous note about using x1, x2, x3 where you are using y1, y2, y3, to avoid later notational confusion; nothing wrong with your method in any case, just a couple of details)*@
3y_2 + 3y_3 = 0
Solution is {(y_1,y_2,y_3) | y_2=s, y_1 = (-1/2)s, y_3 = -s}
eigenvector = r[ (1/2)s; s ;-s] = r[(-1/2);1;-1], so we have y_3 =[(-1/2); 1;-1]*e^(`la*t) = [(1/2); 1;-1] *e^( -2t)
General solution is then
y(t) = c1[0;0;0]*e^(4 t) +c_2 [1; -3/2;0]*e^( t)+ c3[(1/2); 1;-1] *e^( -2t)
y(0) = [-1;0;3], we get
c1[0;0;0]*e^(4 *0) +c_2 [1; -3/2;0]*e^( 0)+ c3[(-1/2); 1;-1] *e^( -2*0), solving
????I believe I meesed up somewhere for one of my values, I’ve gone over it and over it and can’t locate my mistake. I believe the solution is supposed to be….??????
c_1 =0
c_2 = -2
c_3 = -3
??If this were the solution?? Then our solution for init cond y(0) = [-1;0;3] is then
y(t) = (0) [0;0;0]*e^(4 t) +(-2) [1; -3/2;0]*e^( t)+ (-3)[(1/2); 1;-1] *e^( -2t)
= -2e^t [1; -3/2;0] - 3e^( -2t) [(1/2); 1;-1]
So our final solution is
y(t) = -2e^t [1; -3/2;0] - 3e^( -2t) [(1/2); 1;-1]
??????Again this did not exactly work out in my system of equations so this is not for sure the correct, but there were so many places I could have made a simple error that I was not able to locate my mistake???????
@& The matrix A - lambda I for lambda = -2 is [6, 2, 0; 0, 3, 3; 0, 0, 0], so our eigenvector is the solution to the equation
[6, 2, 0; 0, 3, 3; 0, 0, 0] * [x1; x2; x3] = 0
We get the equations
6 x1 + 2 x2 = 0
3 x2 + 3 x3 = 0
In terms of x3 our solutions are
x2 = - x3
x1 = 1/3 x3
so for x3 = 1 our eigenvector would be
[1/3, -1, 1].
Any multiple of this eigenvector would also be valid, and could arise from the same equations.
Thus a solution for lambda = -2 would be
y_1(t) =
@& If we follow the same procedure for eigenvalue lambda = 4 we get the equations
2·x2 = 0
3·x3 - 3·x2 = 0
- 6·x3 = 0
leading us to solution x2 = 0, x3 = 0.
x1 is unaffected and can have any value. Letting x1 = 1 we get eigenvectxor
[1; 0; 0]
Any multiple of this is also a valid eigenvector.
Our solution corresponding to lambda = 4 is therefore
y_2(t) = e^(4 t) [1; 0; 0].*@
e^(-2 t) [ 1/3; -1; 1 ].
*@
@& lambda = 1 leads to the equations
3·x1 + 2·x2 = 0
3·x3 = 0
- 3·x3 = 0
so that x3 = 0 and x1 = -2/3 x2.
Letting x2 = 1 we get eigenvector
[-2/3, 1, 0]
and solution
e^t [ -2/3; 1; 0].*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(Q)’ = flow out + flow in, where flow rate is always r and solute leaving or entering is wrt tank is is leaving or entering from
Q_1 has 3 pipes leaving and 3 pipes entering, but only 2 pipes entering contain solution other contains only fresh water therefore does not contribute to salt contained in tank. Note this is the same for every tank except wrt concentration levels we mult our rate of flow by. Again r is rate of flow Q_Whatever is going to be our salt contained within the tank and V is volume of the tank so we can find the concentration of our solution. Listing all elements out at first we have
(Q_1)’ = -r*(Q_1/V) +(- r*(Q_1/V)) + (-r*(Q_1/V) ) + (r*(Q_2/V)) + (r*(Q_3/V)), so again note we have 3 elements representing out flow(due to 3 pipes flowing solution from tank) from tank and 2 elements representing in flow to tank, where each in flow element is either wrt Q_2 or Q_3. Other tanks Q_2 and Q_3 will follow this same structure of in/out flow
(Q_2)’ = -r*(Q_2/V) +(- r*(Q_2/V)) + (-r*(Q_2/V) ) + (r*(Q_1/V)) + (r*(Q_3/V))
(Q_3)’ = -r*(Q_3/V) +(- r*(Q_3/V)) + (-r*(Q_3/V) ) + (r*(Q_2/V)) + (r*(Q_1/V))
Our system would then be
[Q_1’;Q_2’;Q_3’] = (r/V)[-3,1,1; 1,-3,1; 1,1,-3]*[Q_1;Q_2;Q_3]
???? I had a hard time seeing how to use the notation of systems to put together the system of equations so when looking through book I found where they had already done it. I believe that is where some of my problems are coming from because I understand how to go about finding egienvalues/vectors and finding solutions in other respects. Still have some confusion with substituting in other functions for sets of solutions, like psi(t) in for [y_1,y_2,….,y_n]. All and all though have a fairly good understanding of what’s going on!!?????????
@& You could run into a problem where you are asked to find the matrix as well as solve for the eigenvalues, so review class notes and other notes in the text related to mixing. You should probably first find the system of equations that relate the concentrations in the tanks, then write the system as a matrix.
You do have a good understanding of the solution process once you have the system of equations or the matrix.*@
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
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@& You're doing well with the process. Check my notes on a few details.*@
#$&*
course Mth 279
4/30 3
Query 26 Differential Equations*********************************************
Question: Find the solutions to y ' = A y when
A = [ 5, 3; -4, -3 ]
and
y(1) = [2; 0].
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
(A - `lambda*I)x = 0, going to abbreviate w/ lambda = `la
= [ 5 -`la, 3; -4, -3-`la ]
det[ 5 -`la, 3; -4, -3-`la ] = (5-`la)*(-3-`la) + 12 = [`la^2 -2`la - 15] + 12
= `la^2 -2`la - 3
eigenvalues =`la = -1,3
Eigenvector for `la = -1
[(5-(-1)), 3;-4, (-3-(-1))] *[y_1;y_2] = 0
6y_1 + 3y_2 =0
-4y_1 - 2y_2 = 0
eigenvector = r[1; -2], so we have y_1 =[1; -2]*e^(`la*t) = [1; -2]*e^(- t)
Eigenvector for `la = 3
[(5-(3)), 3;-4, (-3-(3))] *[y_1;y_2] = 0
2y_1 + 3y_2 =0
-4y_1 - 6y_2 = 0
eigenvector = r[1; -2/3], so we have y_1 =[1; -2/3]*e^(`la*t) = [1; -2/3] *e^(3 t)
General solution is then
y(t) = c1[1; -2]*e^(- t) +c_2 [1; -2/3] *e^(3 t)
For y(1) = [2;0], we get
c1[1; -2]*e^(- 1) + [1; -2/3] *e^(3) = [2;0], solving
c_2 =(3/2*e^(3))
c_1 = (e/2)
Solution for init cond y(1) = [2;0] is then
y(t) = (e/2) [1; -2]*e^(- t) + (3/2*e^(3)) [1; -2/3] *e^(3 t)
= (1/2)[1;-2] + (3/2)[1;-2/3]
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
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Self-critique rating:
05-03-2011