Fall 2004 - Holst's Class - Exam 2

# Fall 2004 - Holst's Class - Exam 2 - Name and TA Section...

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Unformatted text preview: Name and TA Section: Student Number: Midterm #2, Math 20D, 2004 Fall Quarter Place/Time: PCYNH/MULTI 106, 9:00-9:50am, 19 November 2004 Instructions: Please write your name and/or student number on each page of the exam, and then solve the following four problems. If you need extra space, use the back of each page of the exam (in this case, clearly indicate which problem you are solving). Problem 1. (25 points) We are given the following IVP: y + ty = t, on (0 , ∞ ) , y (0) = . (a) (20 points) Find the general solution. SOLUTION: This is a linear first-order inhomogenous ODE of the form: a 1 ( t ) y + a ( t ) y = g ( t ) , where a 1 ( t ) = 1, a ( t ) = t , g ( t ) = t . We can use either an integrating factor, or find the homogeneous solution and then use variation of parameters. In either case, we will first write it in the form: y + p ( t ) y = f ( t ) , where p ( t ) = a ( t ) a 1 ( t ) = t, f ( t ) = g ( t ) a 1 ( t ) = t. Integrating factor approach: We first compute the integrating factor m ( t ) = e R p ( t ) dt = e R tdt = e t 2 / 2 . We then have [ e t 2 / 2 y ] = [ my ] = my + mp ( t ) y = mf ( t ) = te t 2 / 2 , so that e t 2 / 2 y = e t 2 / 2 + C, giving y ( t ) = 1 + Ce- t 2 / 2 . Homogeneous solution plus variation of parameters approach: We first solve the homogeneous problem: y h + p ( t ) y h = 0 , which is y h ( t ) = Ce R- p ( t ) dt = Ce R- tdt = Ce- t 2 / 2 . (As usual, we have y h ( t ) = 1 /m ( t ), ignoring the constant.) We then find a particular solution in the form y p ( t ) = y h ( t ) w ( t ). We know that w ( t ) always satisfies the ODE: y h w = f, which comes from simply plugging y p ( t ) into the ODE y + p ( t ) y = f ( t ), and using the fact that y h ( t ) is the homogeneous solution. (Note if we work with a 1 ( t ) y + a ( t ) y = g ( t ), then the ODE w ( t ) satisfies is just slightly different: a 1 y h w = g ; we get exactly the same result for w ( t ).) Now, e- t 2 / 2 w ( t ) = y h ( t ) w ( t ) = f ( t ) = t, giving w = te t 2 / 2 , or w ( t ) = e t 2 / 2 . Finally then y p = y h w = e- t 2 / 2 e t 2 / 2 = 1, and so y ( t ) = y h ( t ) + y p ( t ) = Ce- t 2 / 2 + 1. (b) (5 points) Find the specific solution corresponding to the given initial condition. SOLUTION: We have that: 0 = y (0) = Ce + 1 = C + 1 , so that C =- 1, and thus the solution is y ( t ) = 1- e- t 2 / 2 . 1 Name and TA Section: Student Number: Problem 2. (25 points) Let us assume that the dynamics of the bald eagle population around the world is accurately described by the standard logistic growth with critical threshold model : dy dt =- ry ‡ 1- y T ·‡ 1- y K · , where r > 0 and 0 < T < K . Here, y ( t ) represents the biomass of the eagle population, r represents the ideal reproduction rate without environmental or other limitations,...
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Fall 2004 - Holst's Class - Exam 2 - Name and TA Section...

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