Math 285 Midterm 1 Practice Full Name Net ID UIN Circle your class section B1 C1 F1 You must not communicate with other students during this test No written materials of any kind allowed No phones calculators iPods or electronic devices of any kind allowed for ANY reason including checking the time you may use a simple wristwatch Do not turn this page until instructed to do so There are many di erent versions of this exam Violations of academic integrity in other words cheating will be taken extremely seriously Fill in the following answers on the scantron form now 93 A 94 A 95 D 96 C Multiple Choice Questions Mark answers to questions 1 to 8 on your scantron form 1 3 points The second order di erential equation d2 y dy sin y dt2 dt is equivalent to which pair of first order di erential equations A dw dt dy dt w sin y w y B F w dw dy w sin y C dw dy dy dt w sin y D w dw dt w sin y E dy dt w sin y dy dt w y dy dt dw dy w y w y y w Solution Set w dy dt and use d2 y dt2 d dt dy dt dy dt d dy dy dt wdw dy 2 3 points The Wronskian of the two functions y1 t e3t 2 8 and y2 t e3t 2 4 is A None of the above 2 4 D 12te6t 2 4 E 12te9t 4 12t2 32 B 12te3t C F 0 Solution W e3t 2 8 6te 3t 6te6t 2 2 e3t 2 4 0 e3t 8 3t2 4 4 6te6t 2 4 2 4 6te 3t 0 2 e3t 2 8 0 8 3t2 4 3 3 points The existence and uniqueness theorem for linear di erential equations ensures that the solution of t 2 t 3 y 00 t t 2 y 0 t2 y 1 y 6 2 y 0 6 5 t 5 exists for all t in the region defined by A 1 5 5 3 3 2 2 1 B 2 1 C F 1 5 D 3 1 E 5 3 Solution The points of discontinuity are at t 5 3 2 and t0 6 so the region is the interval 1 5 4 3 points The di erential equation dy t 4y dt 5t y can be transformed by a substitution into the separable di erential equation A 5t y dy t 4y dt B F t dv dt v C dv dt 1 4v 5 v D t dv dt v E t dv dt 1 4v 5 v t 4y 5t y v t2 1 4v 5 v Solution Make the substitution v y t 5 3 points The di erential equation y 0 sin t y cos t y 5 can be transformed by a substitution into the linear equation A v 0 4 sin t v 4 cos t B v 0 2 sin t v C F v 0 2 cos t 4 sin t v 4 cos t D v 0 2 sin t v 2 cos t E v 0 sin t v cos t Solution This is a Bernoulli equation Make the substitution v y 4 1 y4 6 3 points The order of the di erential equation below is higher derivatives are denoted by a Roman numeral i e y iv is d4 y dt4 y 0 5 3 y 0 4 2 y 0 3 A 2 B 5 C F 1 D 4 E 3 Solution 1 because y 0 is the highest derivative that appears y 0 2 10y 0 0 7 3 points Identify which di erential equation below corresponds to the following direction field 6 4 2 0 2 2 A y 0 2 0 2 4 6 y B F y 0 y 3 y C y 0 2 y y D y 0 y E y 0 y y 2 3 3 Solution y 0 y of y 3 y 2 3 y 2 because there are equilibria at y 2 and y 3 and y 2 is stable check the signs 8 3 points For the linear di erential equation t2 y 0 an integrating factor is A e B sin t t cos t 1 cos t C F ecos t D e cos t E cos t Solution e R sin tdt ecos t yt2 sin t t cos t 9 10 points Consider the initial value problem dy dt 3y 4 4t 5 2 y 1 1 a Find the general solution of the di erential equation you do not need to solve for y b Find the unique solution satisfying the initial condition you do need to solve for y this time c On what interval does the solution exist Solution a This problem is separable so we get Z Z dy dt C 2 3y 4 4t 5 2 1 1 3 3y 4 1 1 C 4 4t 5 b Applying the initial condition gives us 1 1 3 3 4 1 1 3 3y 4 1 1 1 1 1 C C 4 4 5 3 4 12 1 1 1 4 3 4t 8 1 4 4t 5 12 3y 4 4t 5 4t 5 4 4t 5 y 1 3 4t 8 c We cannot permit t 5 4 since this is a point of discontinuity of the right side of the di erential equation For any t larger than this the solution exists and the initial value at t 1 is included So the domain is 5 4 1 10 10 points Consider the initial value problem y 00 4y 0 4y 0 y 0 a y 0 0 b a Find a fundamental set of solutions for the di erential equation b What is the general solution of the di erential equation c What is the solution of the initial value problem Solution a The characteristic equation and its solution are r2 4r 4 0 r 2 2 r 2 2 Because of the repeated root a fundamental set of solutions is y1 e2t y2 te2t b The general solution is then y c1 e2t c2 te2t with c1 and c2 arbitrary constants c First compute y 0 2c1 e2t 2c2 te2t c2 e2t Then the initial conditions give us c1 a 2c1 c2 b y ae2t b 2a te2t 11 10 points a A certain species has population level p t modelled by the initial value problem dp dt 3p 2 p 8 p p 0 p0 0 On the phase line below label all critical values equilibrium solutions for this model and the direction of flow between them p 2 0 2 4 6 8 10 b On the axes below draw some representative solution curves including the equilibrium solutions and some solution curves above and below each of them p 10 8 6 4 2 t 0 2 c Under what conditions on p0 will the population go extinct When will extinction actually happen Solution a …
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