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course MTH 279
11/6 NoonProfessor Smith,
I understand how to do number 6, I just don’t have time to type it down as I need to get to class to take an exam. Numbers 1, 3, and 4 I feel comfortable with but I was hoping you could give me some tips if I did anything wrong. Numbers 2 and 5 have me stumped. I did find some material on the LRC circuit in the textbook and I have been trying to work through it but I was wondering if you could give me some tips on solving LRC problems and spring-and-dashpot system problems. Or I would also appreciate it you could refer me to where in the notes you covered these subjects. I have watched all the videos on the module and taken notes but didn’t see anything on those two subjects.
Thank you,
" "1. y” + 5y’ + 6y = 0 has solutions y = e^-3t and y = e^(-3t + 2)
Do these functions form a fundamental set on (-infinity, infinity)?
Condition: Number of functions must equal order of equation
Each function must be a solution to the equation
Let y = e^-3t
y’ = -3e^-3t
y” = 9e^-3t
9e^-3t - 15e^-3t + 6e^-3t = 0, Plugging y into original equation, it is a solution
Let y = e^(-3t+2)
y’ = -3e^(-3t+2)
y” = 9e^(-3t+2)
9e^(-3t+2) - 15e^(-3t+2) + 6e^(-3t+2) = 0, Plugging in y, it is a solution
Both are solutions and the number of functions equals the order of the equation. It does form a fundamental set.
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You haven't established that these functions are linearly independent. To do that you need to evaluate the Wronskian and determine whether or not it is zero.
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2. An unforced LRC circuit has equation:
Q” + RQ’ + Q/(LC) = 0
If C = 0.0072 Farads and L = 0.02 Henries, what is the form of the general solution.
If the circuit is driven by V(t) = 4volts(sin(100(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 the long term.
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This equation is no different in its form than the equation
y '' + b y ' + c y = g(t).
To solve it you find the characteristic equation of the homogeneous equation and solve it, then you apply the technique of either variation of parameters or undetermined coefficients to find your particular solution.
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When you find the characteristic equation you assume a solution of the form e^(r t). In this case the function is Q(t), not y(t), so we let
Q(t) = e^(r t).
Then
Q'' + RQ’ + Q/(LC) = 0
becomes
r^2 e^(rt) + R * r e^(rt) + e^(rt) / (L C) = 0.
The e^(rt) divides out, leaving
r^2 + r * R + 1 / (L C) = 0
The quadratic formula will give you values for r, and you will get your solution to the homogeneous equation in the usual manner.
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To solve the nonhomogeneous equation note that the nonhomogeneous part of the equation
Q'' + RQ’ + Q/(LC) = 4 volts * sin(100 t)
is just a multiple of a sine function. Sine functions typically come about as the result of varoius derivatives of sine and cosine functions, so the a function
4 volts * sin(100 t)
could come out of the left-hand side if Q was a combination of sines and cosines of 100 t.
We therefore assume a particular solution of the form
Q_p (t) = A cos(100 t) + B sin(100 t).
Plugging this into the equation
Q'' + RQ’ + Q/(LC) = 4 volts * sin(100 t)
we obtain an equation we can solve for the parameters A and B.
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3. Solve y” + y’ - 2y = 0 with y(0) = 1 and y’(0) = 1
y = Ae^(rt), y’ = rAe^(rt), y” = r^2Ae^(rt)
r^2 + r - 2 = 0, (r+2)(r-1) = 0, r = 1,-2
y(t) = Ae^(-2t) + Be^t
y’(t) = -2Ae^(-2t) + Be^t
y(0) = Ae^(-2(0)) + Be^(0), A + B = 1
y’(0) = -2Ae^(-2(0)) + Be^(0), -2A + B = 1
A = 0, B = 1
Solution: y(t) = e^t
4. Solve y” + 3y’ + 5y = 0 with y(0) = -1, y’(0) = 1
r^2 + 3r + 5 = 0, r = (-3 + isqrt(11))/2 , (-3 - isqrt(11))/2
y(t) = Ae^(((-3 + isqrt(11))/2)t) + Be^(((-3 - isqrt(11))/2)t)
= Ae^(-3t/2)e^(it(sqrt(11))/2) + Be^(-3t/2)e^(-it(sqrt(11))/2)
= e^(-3t/2)(A(cos(sqrt(11)t/2) + i(sin(sqrt(11)t/2))) + B(cos(sqrt(11)t/2) - i(sin(sqrt(11)t/2)))
y(t) = e^(-3t/2) ((A+B)cos(sqrt(11)t/2) + i(A-B)sin(sqrt(11)t/2))
Forms: y(t) = e^(-3t/2)(c(sub1)cos(sqrt(11)t/2) + c(sub2)sin(sqrt(11)t/2))
Or y(t) = e^(-3t/2) (C(cos((sqrt(11)t/2) + phi)))
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 4m/s at 20 cm from equilibrium, what is its max displacement?
Note: 20 cm on either side, answer both and compare.
Explain behavior of the physical system.
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The force arises from the spring, which exerts force -k x at position x, and the drag force, which exerts force -gamma x ' when velocity is x '.
So, in the absence of a driving force, the net force on the system is -gamma x ' - k x.
Newton's Second Law says that the net force is equal to mass * acceleration. Acceleration being the second derivative of the position function x, our equation is therefore
m x '' = -gamma x ' - k x
so that
m x '' + gamma x' + k x = 0.
Dividing by m we get
x '' + (gamma/m) x ' + (k/m) x = 0
This is solved by assuming position function x(t) = e^(r t).
The equation is critically damped if the discriminant of the resulting characteristic equation is zero. (The discriminant is easily enough seen to be (gamma / m)^2 - 4 k / m. The problem gives you k and m so you can easily find gamma).
Your initial conditions are x ' (0) = 4 m/s and x(0) = 20 cm.
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6. Solve y” - 8y’ + 15y = t^3 with y(0) = 0 and y’(0) =2
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Check the notes I've inserted. I'll be glad to answer followup questions.
I can't give you a reference in the DVD's from here, but if you email me before tomorrow I can give you one when I'm on campus tomorrow.
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Note that the LRC and spring-mass-dashpot systems are of exactly the same form, but with different interpretations. They are solved by exactly the same techniques.
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