#$&* course MTH 279 6/16/13 10:47pmHaven't been able to keep up with work as I intended to because of 2 jobs however next week conditions will have changed and I should be able to complete everything in a timely manner.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Doesn't seem to be very clean, might be in the initial condition but I don't feel very comfortable with the answer ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: 2. t^2 y ' - 9 y = 0, y(1) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = C e^ (- int(9 / t^2 dt)) = C e^ (- (- 9 / t)) = C e^ (9 / t) 2 = C e^(9 / 1) Plug in initial condition 2 = C e^(9) C = 2 / e^(9) y = (2 / e^(9)) e^(9 / t) Plug C into general solution to determine the particular solution confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Same critique as number 1 ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = C e^(- int((2t+1)/(t^2 + t) dt) y = C e^(- ln(t) - ln(t+1)) This solution can not be right, when plugging a 0 in ln(t) you get undefined, what am I missing here???
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 0 ********************************************* Question: 4. y ' + sin(3 t) y = 0, y(0) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = C e^(- int(sin(3t) dt)) y = C e^(cos(3t) / 3) 2 = C e^(cos(3(0)) / 3) Plug in initial condition 2 = C e^(1 / 3) Cos(0) is equal to 1 C = 2 / e^(1 / 3) y = (2 / e^(1 / 3)) e^(cos(3t) / 3) Plug C into general solution to determine the particular solution confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Solution still looks wrong, but I don't think I am doing the math wrong, because when I work the problems from the book I find the right answer ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did. y ' - t^2 y = 0 **** E, the only direction field where the slope changes to zero and then back to what it began as, this is due to the t^2 not be affected by negative numbers so it makes sense the left and right hand side will be the same #$&* y ' - y = 0 **** A, the only direction field where the slope stays the same across the t axis, this is due to the equation not having a t in it #$&* y' - y / t = 0 **** C, because y' = y / t, the slopes will be getting smaller and smaller due to t growing faster than y #$&* y ' - t y = 0 **** B, because when finding the slope, y' = t y, in the first quadrant the slopes will be positive and the fourth quadrant negative #$&* y ' + t y = 0 **** F, because y' = -t y, in the first quadrant the slopes will be negative and the fourth quadrant negative #$&* A B C D E F 6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We will use the rise / run method to find our slope or rather y' (8-2) / (3-1) = 3 = y' plug this into the original 3 = -b y We will use the point (3,8) to solve for b 3 = -b (8) b = - 3 / 8
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am unsure of the answer to this problem as well, thinking about it y' is most likely correct but my approach to finding b is off in my opinion seeing as which ever of the y - coordinates you plug in will change the final answer. ------------------------------------------------ Self-critique rating: 2 ********************************************* Question: 7. The equation y ' - y = 2 is first-order linear, but is not homogeneous. If we let w(t) = y(t) + 2, then: What is w ' ? **** w' = (2+y) + 2 We get 2 + y from y' = 4 + y #$&* What is y(t) in terms of w(t)? **** y(t) = w(t) - 2 #$&* What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ? **** () - w + 2 = 2
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): None ------------------------------------------------ Self-critique rating: 1 ********************************************* Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Looking at the graph it is easy to see that y(0) = y_0 is equal to 1 which is also our C value We can approximate b by using the point (-1, 0.5) y = e^(-b t) 0.5 = e^(-b (-1)) 0.5 = e^(b) take ln of both sides ln(0.5) = ln(e^(b)) b = - 0.69 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Solution is probably wrong but I decided to give it a try ------------------------------------------------ Self-critique rating: ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!