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course MTH 279
8/20 7This is my practice exam. Any tips and pointers would be greatly appreciated.
Thanks!" "1. Solve the equation y ' - 2 y = -1 for y(1) = 2.
dy/dt - 2y = -1
dy/dt = 2y - 1, rearrange
(1/(2y - 1))dy = 1dt, Integrate both sides
(1/2)ln(2y - 1) = t + c, Multiply both sides by 2, and raise to exponent of e
2y - 1 = e^(2t + c), Bring out c from the exponent
2y - 1 = Ce^(2t)
y = Ce^(2t) + ½
y(1) = 2:
2 = Ce^(2(1)) + (1/2)
2 = Ce^2 + (1/2)
1.5 = Ce^2
C = .203
y = .203e^(2t)+ (1/2)
[use integrating factor e^-(2t) to get (y e^(-(2t) ) ' = e^(-2 t) with solution implicit in y e^(-2 t) = 1/2 e^(-2t) + c so that y = 1/2 + c e^(2 t); evaluate c and check that solution works]
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Good.
Note that some problems dictate which method to use or not use. For example you might be asked to solve an equation of this type without using separation of variables, so watch for that.
It's very good that you addressed both possible solution methods here, both of which you have demonstrated on a number of homework problems.
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2. The temperature of a room approaches the outdoor temperature at a rate proportional to the difference between the two temperatures. The outdoor temperature is -10 Celsius, and the initial room temperature is 20 Celsius. The average rate at which temperature changes during the first 30 minutes is -.2 Celsius / minute.
Write the differential equation for this situation, and use it along with the given conditions to find the temperature of the room as a function of clock time.
Do your best to answer the following question for this room: Any reasonable attempt will be given at least partial credit. A complete solution will get some extra credit. An acceptable full-credit answer would include either of the following:
• A correct differential equation which could be solved to answer the question.
• A good graphical description of the direction field of the equation.
[the temperature T approaches room temperature at a rate proportional to the temperature relative to room temperature, so
dT/dt = k (T - (-10 C) ).
Solve this equation, leaving k as an unknown parameter.
Find the temperature after 30 minutes, and use this along with your solution function to evaluate k.
Plot the resulting curve along with your direction field and verify that the curve and the field are consistent.]
dT/dt = k(T - T_r), T = 20 C, T_r = -10 C, T(0) = 20 C
1/(T-T_r)dT = kdt, rearrange and integrate both sides
Ln(T - T_r) = kt + c
T - T_r = Ce^(kt), rearrange
T = T_r + Ce^(kt)
20 = -10 + Ce^(k(0)), plug in given values
C = 30 degrees Celsius
T(t) = -10 + 30e^(kt)
dT/dt = k(T - (-10))
-2 = k(T + 10)
-2 = k((-10 + 30e^(kt)) + 10)
I was unsure how to approach this second part of the problem.
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It's not clear whether you are talking about the direction field or the soft drink problem below. I'll assume the latter.
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A soft drink is taken from the refrigerator at 3 Celsius, and set in the room. If the room had stayed at 20 Celsius, the temperature of the soft drink would have reached 15 Celsius after 40 minutes. What will be the temperature of the room after 40 minutes, and what will be the temperature of the soft drink at this time?
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I believe in this case the full problem would have had the temperature of the room falling or rising at a constant rate.
So the differential equation would be
dT/dt = k ( T(t) - T_room(t))
where T_room(t) is the linear function describing room temperature.
In this case you can still separate variables, but the integration becomes slightly more involved.
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3. Find the exact equation corresponding to dF = 0, where F(t, y) = sqrt(t) * y^3. Explain how you would have solved this equation had you not been given the function F(t, y), and apply the condition y(0) = 1 to find a specific solution.
I understand the given explanation and the process makes sense but I do not know to begin to approach the problem.
[dF = F_x dx + F_y dy, so the equation dF = 0 would be F_x dx + F_y dy = 0. The partial derivatives of the F function comprise the N and M functions of the exact equation. The equation must pass the test for exactness, since F_x y = F_y x for any differentiable function F(x, y).]
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Your note is exactly correct, exept that F is a function of t and y, not x and y.
dF = 0 would be
F_t dt + F_y dy = 0.
This is your equation.
It is automatically so that F_t y = F_y t, but you should in any case verify this.
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4. A swimmer pushes off the side of a pool and glides, slowing to a speed of 10 centimeters /second in 5 seconds after her feet lose contact with the wall, during which time she travels 300 centimeters. Assuming that the net force acting on the swimmer is proportional to her velocity, so that her acceleration is proportional to her velocity, what are her initial velocity and the drag constant. This can be solved without using her mass, but if you wish you may assume a reasonable mass.
F = -kv
Function of time:
m(dv/dt) = -kv
1/vdv = -k/mdt
Ln(v) = -(k/m)t + c
v(t) = Ce^((-k/m)t)
v(t) = Ce^((-k/m)(5))
10 (cm/s) = Ce^((-k/m)(5))
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Good.
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Function of Position:
mvdv/dx = -kv
1dv = (-k/m)dx
v = (-k/m)x + c, c can be interpreted as initial velocity
v(x) = v_0 - (k/m)x
v(300) = v_0 - (k/m)(300)
10 (cm/s) = v_0 - 300(k/m)
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Good.
Between this solution and the preceding you should be able to evaluate your constants, with m remaining unspecified.
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[The force on the swimmer is of the form F = - c v so the m dv/dt = - c v, or m v dv/dx = - c v. These equations can be solved to get v(t), or v(x), as appropriate to the given information.]
5. Find the general solution of the equation y ' + t^2 / y - 4 y = 0.
y’ + t^2/y - 4y = 0
y’ - 4y = -t^2/y
y’ - 4y = -t^2y^-1
Bernoulli: y’ + p(t)y = qy^n
v = y^m, so y = v^(1/m), n = -1, p(t) = -4, q(t) = -t^2
v’ + (1-n)p(t)v = (1-n)q(t)
v’ + (2)(-4)v = (2)(-t^2)
v’ - 8v = -2t^2
e^(-8t)(v’ - 8v) = -2t^2e^(-8t)
Integrate both sides and the right can be solved with integration by parts
[This can be rearranged into the standard form of a Bernoulli equation and solved accordingly.]
6. Solve the equation y ' - 8 sin(2 t) = 6 y.
y’ + p(t)y = g(t)
y’ - 6y = 8sin(2t)
Using integration factor, e^(-6t)
y’e^(-6t) - 6ye^(-6t) = 8sin(2t)e^(-6t)
y’e^(-6t) - (e^(-6t))’y = 8sin(2t)e^(-6t)
(ye^(-6t))’ = 8sin(2t)e^(-6t)
Integrate both sides and the right can be solved with integration by parts
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You need to complete this by integrating both sides and solving for y, but this is straightforward, as it appears you understand.
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[This equation can be rearranged into the form of a linear equation and solved using an integrating factor.]
7. A 1% saline solution flows into a 100-gallon tank initially full of 4% saline solution. A thoroughly mixed solution flows out of the tank at the same rate as the inflow. What is the rate of inflow if after 1 hour the concentration in the tank is 3% saline?
Even having went over the video examples, I am having trouble with this problem. Are there any online examples or pages in the text you would recommend?
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I don't have the text handy here, but there are ample examples and discussion within the first three chapters.
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I just typed 'differential equations mixture problems' into a search engine and got numerous sites, many of which (e.g., Paul's Notes) I have found to be excellent.
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8. Find the solution of the equation y ' = y^2 x / 3 for initial condition y(0) = 1. Evaluate your solution at x = 1.
I am still working through Euler’s examples and getting the hang of it but I am skipping this out of desire to get this submitted before it gets too late so you can review it.
Thanks
Use an Euler approximation with step 1/4 to estimate the solution at x = 1.
Your estimate will either be high or low. Which is it and why is it so?
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I think you're in pretty good shape. Check my notes, and be sure to check out mixture problems. I don't know if they appear on your specific test, but they do appear on about 50% of all versions.
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