#$&* course MTH 279 Section 2.2*********************************************
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: 2. t^2 y ' - 9 y = 0, y(1) = 2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: dy/dt = 9y/t^2 integrate(1/(9y) dy) = integrate(1/t^2 dt) ln(9y)/9 + c_1 = -1/t + c_2 ln(9y) = -9/t + C 9y = Ae^(-9/t) y_g = Ae^(-9/t) 2 = Ae^(-9/1) A = 2e^(9) y_p = 2e^(-9/t + 9) Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (t^2+t)y’ = -(2t+1)y dy/dt = -(2t+1)y/(t^2+t) integrate(dy/y) = integrate(-(2t+1)dt/(t^2+t)) ln(y) + c_1 = -ln(t^2+t)+c_2 ln(y) = -ln(t^2+t) + C y_g = 1/(t^2+t) + e^(C) = A/(t^2+t) This y_g will never fulfill the initial conditions imposed, as a substitution of 0 for t will yield division by 0, which is undefined. There is obviously a vertical asymptote at t = 0. So there is no particular solution to the set of initial conditions given. Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I feel pretty confident that there is no particular solution to this problem. It makes sense to me that there would be an asymptote at t = 0, which, along with division by 0, would make the imposed initial condition false.
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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; I began with t = 0, which would yield a y’ of 0, which means that for every y-value corresponding to t=0 on the t-y plot, the slope would be 0. I narrowed down the choices. I then did the same for t = 1, which yields a y’ that is directly correlated on a 1:1 basis with y. I narrowed down the choices. I incremented t by 1, rationalizing that there is an exponential jump in y’ for every increase in t, as well as a 1:1 correlation as y changes. The limit as t -> inf of y’ yields infinity above the t-axis, and -infinity below the t-axis. This fits E best. **** #$&* y ' - y = 0 A; y’ depends only on the y-value. This means that all slopes in the direction field are positive above the t-axis, and negative below the t-axis, as well as 0 along the t-axis itself. This best fits A. **** #$&* y' - y / t = 0 C; y’ depends directly on y and inversely on t. The more important thing to notice, however, is that when y and t have opposing signs, y’ is negative, and when they have similar signs, y’ is positive. Also, when y and t have the same absolute value, y’ is +/-1, meaning that there are slant asymptotes. When t = 0, y’ is undefined, or vertical. When y = 0, y’ is 0. This best fits C. **** #$&* y ' - t y = 0 B; When t or y are equal to 0, y’ is also equal to 0. There is a direct correlation with both variables, which are multiplied. This means that as t increases, so does y’. As y increases, so does y’. The greatest slopes occur as both y and t increase. Values in the first and third quadrants yield positive slopes, while values in the second and fourth yield negative slopes. This best fits B. **** #$&* y ' + t y = 0 F; y’ is (negatively) directly correlated to the product of t and y. Values in the first and third quadrants yield positive y’ values. Values in the second and fourth quadrants yield negative y’ values. This best fits F. **** #$&* 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: dy/dt = -by integrate(dy/y) = integrate(-bdt) ln(y)= -bt + C y = Ae^(-bt) Imposing (1,2) : 2 = Ae^(-b) b = -ln(2/A) y_g = Ae^(ln(2/A)t) Imposing (3,8): 8 = Ae^(ln(2/A)*3) = A*(8A^(-3)) = 8/A^2 A^2 = 1 Sqrt(A^2) = A = 1 b = -ln(2/A) = -ln(2) y_p = e^(ln(2)t) = 2^(t) y’ - ln(2)y = 0 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: 7. The equation y ' - y = 2 is first-order linear, but is not homogeneous. If we let w(t) = y(t) + 2, then: What is w ' ? **** #$&* What is y(t) in terms of w(t)? **** #$&* What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ? **** #$&* Now solve the equation and check your solution: Solve this new equation in terms of w. **** #$&* Substitute y + 2 for w and get the solution in terms of y. **** #$&* Check to be sure this function is indeed a solution to the equation. **** #$&* YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: w’(t) = y’(t) y(t) = w(t) - 2 w’(t) - w(t) + 2 = 2 >> w’(t) = w(t) integrate(dw/w) = integrate(dt) ln(w) = t + C w = Ae^(t) y + 2 = Ae^(t) y = Ae^(t) - 2 It checks out. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* 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’s evident that y_0 = 1. Using this information, we have: dy/dt = by integrate(dy/y) = integrate(bdt) ln(y) = bt + C y = Ae^(bt) 1 = Ae^(b(0)) A = 1 y_g = e^(bt) It looks as if y(-1) is slightly less than 0.5, so we can use this to estimate our value of b. 0.5 = e^(b*(-1)) b ≈ 0.693 We need the value of y(t) to be less than 0.5, so we increase b, slightly. y = e^((0.75)(-1)) ≈ 0.472 This is pretty close. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Honestly, I feel comfortable up until the finding the b. Since there’s only one (obvious) initial constraint, and it directly relates to the value of A, I could only think to estimate to find a reasonable value of b. Self-critique rating: 3 "