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Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Practice 2 Test
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I submitted my practice 2 test to you last week still haven't gotten feedback on it. I will submit it again through this form.
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1. The equation y '' + 5 y ' + 6 y = 0 has solutions y = e^(-3 t) and y = e^(-3 t + 2). Do these functions form a fundamental set on (-infinity, infinity)?
y’_1 = -3e^(-3t)
y’’_1 = 9e^(-3t)
9e^(-3t) + 5(-3e^(-3t)) + 6(e^(-3t)) = 0
yes
y’_2 = -3e^(-3t+2)
y”_2 = 9e^(-3t+2)
9e^(-3t+2) + 5(-3e^(-3t+2)) + 6(e^(-3t+2)) = 0
yes.
both y1 and y2 are a fundamental set on (0infinity, infinity)
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To test whether this is a fundamenatl set you need to evaluate the Wronskian. It appears that you have added the function and its first two derivatives, which is not related to the Wronskian.
Be sure you have the right definition of the Wronskian. If you do the exercise is pretty straightforward.
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2. An unforced LRC circuit has equation Q '' + R Q ' + Q / ( L C) = 0. If C = .0072 Farads and L = .02 Henries, what is the form of the general solution? If the circuit is driven by voltage V(t) = 4 volts * sin(100 rad/s * t), what is the general solution for which Q(0) = 0 and Q ' (0) = 0? Describe the behavior of the circuit in its transient stage (near t = 0) and in in the long term.
if R = 20000 ohms
Q” + 20000Q’ + 1/(.0072*.02) Q = 0
r^2 + 20000r + 6944.44 = 0
r = -20000 +- sqrt(20000^2 -4(6944.44))/2 = -20000 +- sqrt(4*10^8)/2 = (-20000 +- (2*10^4))/2
so r = 0 and r = -20000
Q(t) = C_1e^(-20000t) + C_2te^(-20000t) general solution
If V(t) = 4sin(100t) Q(0)=0 and Q’(0)=0
not sure how to go about the second part of the question???
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Since I = Q ' , and since an unforced circuit has zero voltage, your equation is entirely equivalent to the given equation Q '' + R Q ' + Q / (L C) = 0.
You could solve using the equation you state, which is a first-order nonhomogeneous linear equation with constant coefficients, and it wouldn't be a bad exercise to solve it this way.
However you will probably find it easier, and more pertinent to this chapter, to solve the equation in terms of Q. The equation would be
Q '' + R Q ' + Q / (L C) = V(t).
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Your solution will consist of the solution to the homogeneous equation, plus a solution to the nonhomogeneous equation.
As t approaches infinity the solution will become stable and repetitive.
Until it does to, the system is said to be in the transient state. So you'll need to describe the behavior of the system for some finite time after the initial instant.
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Do you use the equation V(t) = RI + LI’ + Q/C???
If so, do you plug in the given values and solve for I???
3. Solve the equation y '' + y ' - 2 y = 0 with y(0) = 1 and y ' (0) = 1.
r^2 + r - 2 = 0
(r +2)(r-1) = 0
r =-2 r=1
y(t) = Ae^(-2t) + Be^t general solution
y(0) = Ae^0 + Be^0 = 1
A + B = 1
y’ = -2Ae^(-2t) + Be^t
y’(0) = -2Ae^0 + Be^0 = 1
-2A + B = 1
solve system
A + B = 1
-2A + B = 1 (subtract)
3A = 0
A = 0
B=1
y(t) = e^t
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You should plug this back into the equation to verify that it is a solution, and also plug in the initial conditions to ensure that the solution meets them.
For this particular solution, it is very easy to verify that it does work.
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4. Solve the equation y '' + 3 y ' + 5 y = 0 with y(0) = -1 and y ' (0) = 1.
r^2 + 3r + 5 = 0
r = -3 +- sqrt(3^2 -4(5))/2 = -3 +- sqrt(-11)/2
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r = -3/2 +- sqrt(11)/2 * i.
Your next step does follow from this corrected version of the solution, so I suspect you just neglected to type in the i.
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y(t) = e^(-3/2t)(Acos(sqrt(11)/2*t) + Bsin(sqrt(11)/2*t))
y(0) = e^0(Acos(0) + Bsin(0)) = -1
-1 = A
y’(t) = -3/2 *e^(-3/2*t)(sqrt(11)/2*Asin(sqrt(11)/2 * t) + sqrt(11)/2 * Bcos(-sqrt(11)/2 * t)
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You have not used the product rule correctly in calculating the derivative.
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y’(0) = -3/2(sqrt(11)/2 A sin(0) + sqrt(11)/2 Bcos(0)) = 1
1 = -3/2(sqrt(11)/2 Bcos(0))
-2/3 = sqrt(11)/2 B
B = -4/(3sqrt(11))
Not really sure about this B value???
y(t) = e^(-3/2*t)(-1cos(sqrt(11)/2*t) + (-4/(3sqrt(11))sin(sqrt(11)/2 * t)
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You need to correct your calculation of the derivative and use it as a basis for your solution.
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5. A spring-and-dashpot system has mass 2 kg and force constant 800 N / m. For what drag constant is this system critically damped? If the critically damped system is given an initial velocity of 4 m/s at a position 20 cm from equilibrium, what is its maximum displacement from equilibrium? Note that a position 20 cm from equilibrium could be on either side of equilibrium; answer for both caes, and compare your results. Explain the comparison in terms of the behavior of the physical system.
A) is the drag constant gamma in the following equation:???
mx” + gamma x’ + kx = 0
if so for critically damped
gamma^2 = 4km
gamma^2 = 4(800)(2)
gamma = +- 80 ???
B? v = 4m/s x=20cm
find max displacement???
I am not sure what equation to use and how to solve this problem???
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The equation is m x '' + gamma x ' + k x = 0.
gamma can't be negative so your result will be gamma = 80.
gamma also has units, which should make the term gamma x ' consistent with the units of the other terms of the equation. It isn't absolutely necessary to use the units, but it's a very good idea if you know how because it helps greatly to ensure that your solution is consistent.
In any case your equation (omitting units for brevity) is
2 x '' + 80 x ' + 800 x = 0.
x is the position of the mass so x ' is its velocity and x '' is its acceleration.
You don't have to know this here but you might for other problems: The equation simply expresses Newton's Second Law, which is
m x '' = F_net.
The net force is - 80 x ' - 800 x.
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6. Solve the equation y '' - 8 y ' + 15 y = t^3 with y(0) = 0 and y ' (0) = 2.
y_c?
r^2 -8r + 15 = 0
(r-5)(r-3) = 0
r=5 r=3
y_c = C_1e^(5t) + C_2e^(3t)
In general do it matter which r value goes with C_1 and which goes with C_2 or will the problem and math work itself to the right answer no matter which order you put the r’s???
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The equation can be factored in either order, so there's no need to worry about the order of your terms. The mathematics will, as you say, sort itself out.
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y_p?
g(t)=t^3
use the general form t^r[At^n + Bt^n-1 +…]
here r =0
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Right. Your nonhomogeneous expression does not involve one of the functions of your solutoin y_c.
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y_p = At^3 + Bt^2 + Ct + D
y’ = 3At^2 + 2Bt + C
y” = 6At + 2B
plugging in
6At + 2B -8(3At^2 + 2Bt + C) + 15(At^3 + Bt^2 + Ct + D) =t^3
factoring thru you get
6At + 2B -24At^2 - 16Bt - 8C + 15At^3 + 15Bt^2 + 15Ct + 15D =t^3
creating system
15A = 1
-24A + 15B = 0
6A -16B + 15C = 0
2B -8C + 15D = 0
A = 1/15
-24(1/15) + 15B = 0
15B = 8/5
B = 8/75
6(1/15) - 16(8/75) + 15C = 0
-98/75 + 15C = 0
15C = 98/75
C = 98/1125
2(8/75) - 8(98/1125) + 15D = 0
15D = 1024/1125
D = 544/16875
putting it all together
y(t) = C_1e^(5t) + C_2e^(3t) + t^3/15 + 8/75t^2 + 98/1125t + 544/16875
solving for C1 and C2
y(0) = C1 + C2 + 544/16875 = 0
C1 + C2 = -544/16875
y’ = 5C_1e^(5t) + 3C_2e^(3t) + 3/15t^2 + 16/75t + 98/1125
y’(0) = 5C_1 + 3C_2 + 98/1125 =2
5C_1 + 3C_2 = 2152/1125
C_2 = -544/16875 - C_1
5C_1 -3(-54/16875 - C_1) = 2152/1125
5C_1 - 3C_1 - 544/5625 = 2152/1125
2C_1 = 1256/625
C_1 = 628/625
C_2 = -28/27
final answer
y(t) = 628/625*e^(5t) + (-28/27)e^(3t) + t^3/15 + 8/75t^2 + 98/1125t + 544/16875
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i haven't checked all the arithmetic in your solution, nor have I checked your solution in the equation and initial conditions, but your procedure is valid throughout. If you did the arithmetic and algebra right, your solution should check out.
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You're not in bad shape here, but you are missing a few things so check my notes out carefully.
I've emailed your Test 2 to your proctor. I got a reply that he is out of town until next week, and resent the tests to the address of the center. Check with them to be sure they have what they need, and let me know if they don't.
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