APPM 2360: Midterm exam 1 February 15, 2017

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1 APPM 36: Midterm exam 1 February 15, 17 On the front of your Bluebook write: (1) your name, () your instructor s name, (3) your recitation section number and () a grading table. Text books, class notes, cell phones and calculators are NOT permitted. A one page (letter sized 1 side only) crib sheet is allowed. Problem 1: [3 pts, 5 pts each] Consider the statements below and determine whether they are True or False. No abbreviation. Box a unique answer for each statement. No justification is required. (a) The ordinary differential equation (ODE) y + t ln(y t ) = t sin y is separable. (b) The conditions of Picard s Theorem are satisfied by the IVP y = y sin t with y(1) =. (c) The conditions of Picard s Theorem are satisfied by the IVP y = y 1 5 cos t with y() = 1. (d) y = 3t + t is a solution to /dt = 1 t y + t, t >. (e) If y 1 and y are both solutions to y + y y =, then so is y 1 + y. (f) A radioactive element has a half-life T 1/. After a time 3 T 1/, one sixth of the initial amount has not undergone radioactive decay yet. (a) True. Since y +t ln(y t ) = t sin y may be rewritten y = t ln(y )+t sin y = t [ sin y ln ( y )]. (b) False. f(t, y) = y sin t is continuous at (, ), but f y (t, y) = 1 y 1 sin t is undefined at (1, ). Hence the conditions of Picard s theorem are not verified for Initial Condition (t, y ) with t = 1 and y =. (c) True. f(t, y) = y 1 5 cos t and f y (t, y) = 1 5 y 5 cos t are both continuous at (t, y ) = (, 1). (d) True. dt = d dt (3t + t ) = 3 + t and 1 t y + t = 1 t (3t + t ) + t = 3 + t. (e) False. Only true for linear homogeneous equations, but y + y y = is non linear. (f) False. After 3 T 1/, the amount of material that has not decayed yet is (1/) 3 = 1/8.

2 Problem : [3 pts] Consider the differential equation ty = t y t e t (1) (a) [6 pts] Classify the differential equation (1) in terms of order, linearity, and homogeneity. (b) [8 pts] Find y h (t), the solution to the homogeneous equation associated to equation (1). (c) [1 pts] Using the method of Variation of Parameters (Euler-Lagrange), find a particular solution y p (t) to equation (1). Show all work: points are given for method. (d) [6 pts] Find and verify the general solution to Eq. (1). (a) This is a first order, linear, nonhomogeneous differential equation. (b) The associated homogeneous equation is ty h = t ty h. Separating gives the equation 1 y h h = tdt. Thus, separation of variables yields y h = ce t. (c) Variation of parameters will produce a particular solution. The normalized form of this differential equation is y = ty te t. The particular solution is given by y p = v (t) e t, where v (t) = te t dt = 1 e t. Thus, y p = 1 e t. (d) The general solution is given by Verification: on one hand: on the other hand: y = y h + y p = ce t + 1 e t. ty = t ce t t e t, t y t e t = t ce t + t e t t e t = t ce t t e t so the general solution satisfies Eq. (1).

3 Problem 3: [8 pts] Consider the initial value problem dt = y + t, y() = 1. (a) [16 pts] Starting at y() = 1 and using a timestep equal to h =.5, compute the first two steps of Euler s method on this equation to obtain an approximation to the solution at t = 1. Note this value y Euler (1). (b) [ pts] It appears that the initial value problem () can be solved exactly. The homogeneous solution is y h = Ke t and a particular solution is y p = (1 + t). What is the solution to the initial value problem? (c) [ pts] Compute the global error of the method after two timesteps: E = y exact (1) y Euler (1). You may use the approximation e 5.. Do not be disappointed if this error is large: it appears that Euler s method with a time step.5 is not very accurate here. (d) [ pts] Suggest two different ways to get a more accurate estimate. (a) y Euler (1) =.5. Method: n t n y n y = y n + t n y n+1 = y n +.5 y = 1 y 1 = = = y = = (b) By superposition, the general solution is y g = Ke t (1 + t). Imposing the initial value y() = 1 sets K = =. The solution of the initial value problem is: (c) The global error is: y(t) = e t (1 + t) E = e 1 (1 + 1) y Euler (1) = 5..5 =.9. The method gives a mediocre result. (d) Euler s method with a smaller timestep would yield a more accurate estimate. So would a higher order method. ()

4 Problem :[3 pts] A tank has capacity 15 litres and is initially filled with 5 litres of pure water in it. You pour a solution with concentration 3 grams of salt per litre into the tank at a rate of litres per minute and drain the well-mixed solution out of the tank at a rate of 1 litre per minute. Show all work: points are given for method. (a) [7 pts] Set up the IVP that describes the amount of salt in the tank after t minutes. (b) [18 pts] Using the method of your choice, solve this IVP. (c) [ pts] What is the concentration of the solution in the tank after t minutes? (d) [3 pts] What is the concentration of the solution in the tank at the exact moment the tank is full? (a) The volume of fluid in the tank is : (b) V (t) = (F in F out ) t + V =. Let x(t) represent the number of pounds of salt in the tank at time t measured in minutes. The IVP to solve is dt = F inc in x(t) V (t) F out with the initial value: dt = 3 x 1 dt = 6 x x() = Using separation of variables to first find x h, the general solution of dt + x = or dt = x we obtain x h (t) = C, C R. To find x p, a particular solution to the nonhomogeneous diffeq, using variation of parameters we try a solution of the form x p (t) = v(t) = x p(t) = v (t) v(t) () (product rule) and substituting this in to the original diffeq yields v (t) v(t) () + v(t) () = 6 so that we may use v(t) = 3(). Thus 3() x p (t) = and the solution to the original diffeq is v (t) = 6() = 3() x(t) = x p (t) + x h (t) = 3() + C.

5 Using the initial condition x() = we find that so the solution to the IVP is Or, using the integrating factors: µ(t) = e = 3( + 5) + C + 5 C = 75 x(t) = 3() t+5 dt = e ln t+5 = where we don t need a +C when finding µ and we may drop the absolute value since >. Multiply both sides of the diffeq by µ(t): () + x = 6() dt d dt [xµ] = 6() () dt d [xµ] dt = 6 dt xµ = 3 () + C, with C R x(t) = 3 () + C Using the initial condition x() = we find that = 3( + 5) + C + 5 C = 75 so the solution to the IVP is (c) The concentration is: x(t) = 3 () 75 C(t) = x(t) V (t) = 3 75 () (d) The tank will be full when V (t) = 15, which corresponds to t = 1 minutes. evaluate: C(1) = = = 8 3 g.l 1 We can

6 Problem 5: [3 pts] These two short independent problem concern graphical methods. (a) [15 pts] Match each of the equations (3a), (3b), (3c), and (3d), below: dt = y, (3a) dt = t + y, (3b) dt = y, (3c) dt = t, (3d) with one of these five direction fields I, II, III, IV, V. One direction field corresponds to none of the equations above: which one? (b) [15 pts] Draw the phase line corresponding to the equation dt = y y 6 () If you find any, characterize the equilibrium solutions: give their position and stability. Without computing the solution but using the phase line to justify your answer, determine the long term behaviour of the solution to the equation with the initial value (t, y ) = (, ). (a) Equation 3a and 3c are the only autonomous equations: they correspond to horizontally invariant direction fields: I, II or IV. The solution to equation 3a is always increasing: it corresponds to I. The solution to equation 3c is decreasing for y > and increasing for y < : it matches IV. The direction field II has no matching equation. The slope to solution 3d does not depend on time: it is associated with III, the only vertically invariant direction field. The solution to equation 3b varies with both y and t, corresponding to V.. (b) This autonomous equation is of the form y = f(y) with: f(y) = y y 6 We construct the phase line from the properties of f. f is defined for all y R, hence the phase line contains no barrier. The equilibria correspond to the zeroes f(y eq ) = : y eq = 1 ± 1 + = or 3 Since f is a concave up parabola: f(y) < for y (, 3), and f(y) > for y < or y > 3. Hence y 1 = is a stable equilibrium point and y = 3 is an unstable equilibrium point. We summarize these findings on the phase line below. The initial condition y =

7 lies in the interval [, 3), and is therefore attracted to the stable equilibrium y 1 =. Hence the limit of the solution to the IVP: lim y(t) = (5) t

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