MATH 23 EXAM 1 REVIEW PROBLEMS

Transcription

1 MATH 3 EXAM 1 REVIEW PROBLEMS Problem 1. A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time. Solution. Let V(t) and S(t) be the volume and surface area, respectivel, of the raindrop at time t. Then V(t) = 4 3 πr(t)3 S(t) = 4πr(t) where r(t) is the radius of the drop at time t. Since the rate of evaporation is proportional to the surface area, then dv dt = αs for some constant α. We now express S in terms of V. Observe that S 3/ = (4πr ) 3/ = 8π 3/ r 3 so S 3/ V = 8π 3/ r 3 = 6π 1/. 4 3 πr3 Then S 3/ = 6π 1/ V = S = 6 /3 π 1/3 V /3 = (36π) 1/3 V /3 so dv dt = αs = α(36π)1/3 V /3. (You are not required to solve this equation in this problem, but ou can using separation of variables, which gives V = (α(36π) 1/3 t + V 1/3 0 ) 3 where V 0 is the initial volume.) Problem. For large, rapidl falling objects, the drag force is approximatel proportional to the square of the velocit. (a) Write a differential equation for the velocit of a falling object of mass m if the magnitude of the drag force is proportional to the square of the velocit and its direction is opposite to that of the velocit. (b) Determine the limiting velocit after a long time. (c) If m = 10 kg, find the drag coefficient so that the limiting velocit is 49 m/s. (d) Using the data in part (c), draw a direction field for the differential equation. Solution. (a) We orient our axes so that down is the positive direction. Then the two forces acting on the object are the force due to gravit, which is mg (where g 9.81m/s is the acceleration due to gravit) and the force due to drag, which is kv for some positive constant k. B Newton s second law of motion, then ma = mv = mg kv 1

2 where a is the object s acceleration. Thus our equation can be written v = g (k/m)v. (b) We seek to find the equilibrium solution, i.e., a function v(t) that makes v (t) identicall zero. We find 0 = v = g (k/m)v = (k/m)v = g = v = mg mg = v = k k. Thus after a long time, we expect the object to have velocit mg/k. (c) (d) 49 = 10g/k = 49 k = 10g = k = 10g Problem 3. According to Newton s law of cooling, the temperature u(t) of an object satisfies the differential equation du = k(u T) dt where T is the constant ambient temperature and k is a positive constant. Suppose that the initial temperature of the object is u(0) = u 0. (a) Find the temperature of the object at an time t. (b) Let τ be the time at which the initial temperature difference u 0 T has been reduced b half. Find the relation between k and τ. Solution. (a) Separating variables, we have du dt = k(u T) = ln u T = du u T = k dt = kt + C 0.

3 1 1:03 c01 Sheet number 18 Page number 18 can black Exponentiating, then u T = Ce kt = u = T + Ce kt Chapter. 1. Introduction The initial condition implies (b) Let τ be the time atuwhich 0 = u(0) the initial = T + temperature C = C difference = u 0 Tu 0 T has been reduced b half. Find the relation between k and τ. 16. Suppose which that gives a building a final loses answer heat of in accordance with Newton s law of cooling (see Problem 15) and that the rate constant k has u the = T value + (u h T)e 1.Assume kt that the interior temperature is 70 F when the heating sstem fails. If the external temperature is 10 F, how long will it (b) take B for the the definition interior temperature of τ, we have to fall to 3 F? 117. (u Consider an electric circuit containing 0 T) = u(τ) T = (u 0 T)e kτ a capacitor, = 1 = resistor, and batter; e kτ = e kτ see Figure The charge Q(t) on the capacitor satisfies the equation 5 = = kτ = ln(). Problem 4. Consider an electric circuit containing a capacitor, resistor, and batter; see R dq + Q the figure below. The charge Q(t) on thedtcapacitor C = V, satisfies the equation where R is the resistance, C is the capacitance, R dq dt + Q and C = V V, is the constant voltage supplied b the batter. (a) If Q(0) = 0, find Q(t) at an time t, and sketch the graph of Q versus t. where R is the resistance, C is the capacitance, and V is the constant voltage supplied b the batter. (b) Find the limiting value Q L that Q(t) approaches after a long time. (c) Suppose that Q(t (a) If Q(0) = 0, find Q(t) 1 ) = Q at L and that at time t = t an time t, and sketch 1 the batter is removed and the circuit the graph of Q versus r. is closed again. Find Q(t) for t > t (b) Find the limiting value Q L that 1 and sketch its graph. Q(t) approaches after a long time. R V C FIGURE 1..3 The electric circuit of Problem 17. Solution. (a) Separating variables, we have 18. A pond containing dq 1,000,000 gal of water is initiall free of a certain undesirable chemical (see Problem dt = 1 ( 1 of Section V Q ) 1.1). = 1 Water containing (CV Q) R C RC 0.01 g/gal of the chemical flows into the pond at a rate of 300 gal/h, and water also flows out of the pond at the same rate. Assume dq that the chemical is uniforml distributed throughout the pond. 1 = ln CV Q = (a) Let Q(t) be the amount of the chemical CV in the Q = pond at C dt = 1 time t. Write RC t + C 0. down an initial value Exponentiating problem for Q(t). both sides ields (b) Solve the problem in part (a) for Q(t). How much chemical is in the pond after 1 ear? (c) At the end of CV 1 ear Q the = source C 1 e RC 1 of t the = chemical Q = CV in the C pond 1 e RC 1 is t removed;. thereafter pure Thewater initial flows condition into the pond, implies and the mixture flows out at the same rate as before. Write down the initial value problem that describes this new situation. (d) Solve the initial value 0 = problem Q(0) = incv part (c). C 1 How = much C 1 = chemical CV remains in the pond after so we 1 additional get a final ear answer ( ears offrom the beginning of the problem)? (e) How long does it take for Q(t) to be reduced to 10 g? (f) Plot Q(t) versus tq(t) for 3= ears. CV CVe RC 1 t = CV(1 e RC 1 t ) Your swimming pool containing 60,000 gal of water has been contaminated b 5 kg of a nontoxic de that leaves a swimmer s skin an unattractive green. The pool s filtering sstem can take water from the pool, remove the de, and return the water to the pool at a flow rate of 00 gal/min.

4 (b) Setting dq/dt = 0, we find 0 = R 0 = V Q L C where Q L is the limiting value. = Q L = CV Problem 5. Find the solution to the initial value problem + = te t, (1) = 0. Solution. The integrating factor is µ(t) = e t. Multipling both sides of the equation b µ, we have d dt [et ] = t. Integrating both sides of the equation results in the general solution (t) = (1/)t e t + ce t. The initial condition implies that (1/)e + ce = 0, hence c = 1/, so the solution of the IVP is (t) = 1 (t 1)e t. Problem 6. Consider the initial value problem + 1 = cos(t), (0) = 1. Find the coordinates of the first local maximum point of the solution for t > 0. Solution. The integrating factor is Therefore the general solution is µ(t) = e (1/)dt = e t/. (t) = 1 5 (4 cos(t) + 8 sin(t)) + ce t/. The initial condition gives a specific solution Differentiating, we find (t) = 1 5 (4 cos(t) + 8 sin(t) 9e t/ ). (t) = 1 ( 4 sin(t) + 8 cos(t) + (9/)e t/ 5 (t) = 1 5 ( 4 cos(t) 8 sin(t) (9/4)e t/. Setting (t) = 0, the first critical value is t Since (t 1 ) < 0, this is a local maximum with approximate coordinates (1.3643, ). Problem 7. Consider the initial value problem 3 = 3t + et, (0) = 0. 4

5 Find the value of 0 that separates solutions that grow positivel as t from those that grow negativel. How does the solution that corresponds to this critical value of 0 behave as t? Solution. The integrating factor is µ(t) = e 3t/. Multipling through, we have d dt [e 3t/ ] = 3te 3t/ + e t/. Integrating, the general solution is The initial condition implies (t) = t 4/3 4e t + ce 3t/. (t) = t 4/3 4e t + ( /3)e 3t/. Now, as t the ( /3)e 3t/ term will dominate, hence its sign will determine whether the solution grows positivel or negativel. Thus the critical value of the initial condition is 0 = 16/3. The corresponding solution decreases without bound. (t) = t 4/3 4e t Problem 8. Consider the initial value problem = x(x + 1) 4 3, (0) = 1. (a) Solve the IVP. (b) Determine the interval on which the solution is valid. Solution. (a) Rewriting the differential equation as 4 3 d = x(x + 1)dx, integration ields 4 = 1 4 (x + 1) + c. The initial condition implies c = 0, which gives an implicit solution (x + 1) 4 4 = 0. The explicit form is then x + 1 (x) = where we have chosen the sign so that (0) = 1/. (b) Since x for all x, then the solution is valid for all x, so (, ) is the interval of validit. Problem 9. Consider the initial value problem t(4 ) =, (0) = > 0. t (a) Determine how the solution behaves as t. (b) If 0 =, find the approximate time T at which the solution first reaches the value

6 Solution. (a) Separating variables, we have d (4 ) = t 1 + t dt. The partial fraction decomposition of the LHS is 1 (4 ) = 1/4 1/4 4 so we have ln ( 1 4 = ln ln 4 = 1 ) d = 4 4 Exponentiating ields We solve for : e 4t e 4t 4 = c 1 (1 + t) 4. = ( 4)c 1 (1 + t) 4 = c 1 (1 + t) 4 4c 1 (1 + t) 4 e = (1 4t ) e c 1 (1 + t) 4 = 4t 4c 1 (1 + t) 4 e 4t e 4t t dt = 4t 4 ln 1 + t + c. 1 + t e4t (1+t) = = 4c 4 4c 1 = 1 e 4t e 1 c 4t 1 (1 + t) 4 c 1 e 4t = 4c 1 e 4t c 1 e 4t (1 + t) 4. (1+t) 4 Thus as t, (t) = 4c 1 e 4t c 1 e 4t (1 + t) 4 = 4 1 (1+t)4 c 1 e 4t (b) The initial condition (0) = implies = 4. = (0) = 4c 1 c 1 1 = c 1 = 4c 1 = = c 1 = c 1 = 1 so 4e (t) = 4t e 4t (1 + t) 4 = 4e 4t e 4t + (1 + t) 4. Using a computer, we find the numerical answer to is t = (t) = 4e 4t e 4t + (1 + t) 4 Problem 10. Using a theorem proved in class, determine an interval on which the solution of the following initial value problem is guaranteed to exist. Be sure to state how ou are using the theorem. (ln(t)) + = cot(t), () = 3 6

7 Solution. Rewriting the equation in standard form, we have + 1 ln(t) = cos(t) sin(t). Since ln(1) = 0 and sin(t) = 0 for integer multiples of π, then the coefficient functions are discontinuous there. The interval containing the initial value t 0 = is (1, π), which is the desired interval. Problem 11. State where in the t, -plane the hpotheses of Theorem.4. are satisfied for the following ODE. d dt = (cot(t)). 1 + Solution. The function f (t, ) is discontinuous along the lines t = ±kπ for k = 0, 1,,... and = 1. The partial derivative f = cot(t) (1 + ) has the same discontinuities. Thus the hpotheses are verified for all points in the t, - plane not mentioned above. Problem 1. Heat transfer from a bod to its surrounds b radiation, based on the Stefan- Boltzmann law, is described b the differential equation du dt = α(u4 T 4 ), where u(t) is the absolute temperature of the bod at time t, T is the absolute temperature of the surroundings, and α is a constant. However if u is much larger than T, the solutions to the above equation are well approximated b solutions of the simpler equation du dt = αu4. (1) Suppose that a bod with initial temperature 000 K is surrounded b a medium with temperature 300 K and that α = K 3 s. (a) Determine the temperature of the bod at an time t b solving (1). (b) Describe the solution s behavior as t. Does this behavior make sense? Solution. The equation is separable and has implicit solution u 3 = With the given values, it follows that u 3 0 3αu 3 0 t + 1. u(t) = t

8 Up to now, we have been primaril concerned with showing that first order differential equations can be used to investigate man different kinds of problems in the natural sciences, and with presenting methods of solving such equations if the are Chapter. First Order Differential Equations (a) As t, we see that u(t) 0. This doesn t make sense: we expect the bod s temperature to approach the ambient temperature of 300 K. This is because we Correct didn tsolutions use the true werestefan-boltzmann found b Johann Bernoulli law and his brother Jakob Bernoulli and b Isaac Newton, Gottfried Leibniz, du dt = and the Marquis de L Hôpital. The brachistochrone problem is important in the development α(u4 of mathematics T 4 ). as one of the forerunners of the calculus of variations. ProblemIn 13. solving In this this problem, problem, we it is will convenient attemptto totake findthe theorigin curveasalong the upper which point a particle P and towill slide without orient thefriction axes asin shown the minimum Figure.3.6. timethe from lower a given point point Q haspcoordinates to another(xpoint 0, 0 ).Itis Q that is lower then than possible P. to show that the curve of minimum time is given b a function = φ(x) that It issatisfies convenient the differential to take the equation upper point P as the origin, and to orient the axes as shown below. The lower point Q has coordinates (x 0, 0 ). One can show that the desired curve is given b a function (t) satisfing the (1 differential + ) = k, equation (i) where k is a certain positive constant (1 + to ( be ) ) determined = k later. where (a) k is Solve some Eq. positive (i) for constant.. Wh is it necessar to choose the positive square root? (b) Introduce the new variable t b (a) Solve the above equation for the relation. Wh is it necessar to choose the positive square root? = k sin t. (ii) (b) Introduce the new variable t given b the equation Show that the equation found in part = (a) k then sin takes (t). the form Show that the equation found in kpart sin (a) tdt= then dx. takes the form k sin (t) dt = dx. () (c) Letting θ = t,show that the solution of Eq. (iii) for which x = 0 when = 0 is given b (c) Letting θ = t, show that the solution of () for which x = 0 when = 0 is given b x = k (θ sin θ)/, = k (1 cos θ)/. (iv) x(θ) = k (θ sinθ), (θ) = k (1 cosθ) Equations (iv) are parametric. equations of the solution of Eq. (i) that passes through (0, (The 0). The graph graph of of theeqs. equations (iv) is called (x(θ), a ccloid. (θ)) is called a ccloid.) (d) (d) If we If we makea a proper choice of the constant k, k, then then the ccloid ccloid also also passes passes through through the point the point (x 0, 0 )(x and 0, is 0 the ) and solution is theofsolution brachistochrone of the problem. described Find k if x 0 = at1the andbeginning 0 =. of the exercise. Find k if x 0 = 1 and 0 =. P x (iii) Q(x 0, 0 ) Solution. FIGURE.3.6 The brachistochrone. (a) Solving the equation for, we find k = 1 = k. 8 Differences Between Linear and Nonlinear Equations

9 We choose the positive root because of how we have oriented our axes: is a nonnegative, increasing function of x. (b) Letting = k sin (t), then d = k sin(t) cos(t) dt. Note that k = k k sin (t) k sin (t) = cos (t) sin (t). Substituting these two expressions into the result of part (a), we have cos(t) sin(t) = k = d dx = d = k sin(t) cos(t) dt dx hence k sin (t) dt = dx. (c) Setting θ = t, then dθ = dt, so we obtain k sin (θ/) dθ = dx. Integrating both sides and noting that t = θ = 0 corresponds to the origin, we find x(θ) = k (θ sin(θ)). To find, we recall the expression from part (b): = k sin (t) = k 1 cos(t) = k (1 cos(θ)). b the half-angle formula. (d) Observe that x = 1 cos(θ) θ sin(θ). Setting x = 1, =, we find the approximate solution of = 1 cos(θ) θ sin(θ) is θ Substituting this into the equation for x(θ) and solving for k ields an approximate solution of k.193. Problem 14. Consider the autonomous ODE below. Let f () = d/dt. d dt = (4 ) < 0 < (a) Sketch the graph of f () vs. b hand (i.e., without using a graphing utilit, such as a graphing calculator). (b) Determine the equilibria, and classif each one as stable, unstable, or semistable. (c) Draw the phase line, and sketch several graphs of solutions in the t, -plane. Solution. (a) Since f () = ( )( + ) = 4 4, then f has zeroes at 0, ±. We compute f () = = 4( ) f () =

10 Choosing test points, we have the following sign chart for f () f () and the following sign chart for f (). f () + /3 /3 With this information about where the f is increasing and decreasing and its concavit, we can make a reasonabl accurate plot of f. 4 f () (b) As we remarked above, f () = d/dt has zeroes at =, 0, which are the equilibria. From the phase line sketch, we see that = is unstable, = 0 is semistable and = is stable. (c) 10

11 Problem 15. Consider the autonomous ODE below. Let f () = d/dt. d dt = (1 ) < 0 < (a) Sketch the graph of f () vs. b hand (i.e., without using a graphing utilit, such as a graphing calculator). (b) Determine the equilibria, and classif each one as stable, unstable, or semistable. (c) Draw the phase line, and sketch several graphs of solutions in the t, -plane. Solution. (a) Using the same steps as outlined in the previous problem, we construct the following plot. f () = d/dt (b) The equilibria = 0 and = 1 are both semistable, as can be seen from the phase line. 11

12 Problem 16. Determine whether the following equation is exact. If it is exact, find the solution. (x + 3) dx + ( ) d = 0. Solution. Since M = 0 = N x, then the equation is exact. Then there is some function ψ(x, ) such that ψ x = M and ψ = N. Then ψ = (x + 3) dx = x + 3x + C() so ψ = C (). Comparing this with N, we find C () = N =, so C() = + C 0. Taking C 0 = 0, then ψ = x + 3x +, so the equation can be written as which has solution x + 3x + = c. d dx ψ = d dx (x + 3x + ) = 0 Problem 17. Find an integrating factor and solve the following equation. + (x e ) = 0. Solution. M = 1 and N x =, so we take as integrating factor ( ) ( Nx ) M d µ() = exp M d = exp = e ln() =. Multipling through b µ, we obtain dx + (x e ) d = 0. Redefining M and N as the coefficient functions of this new equation, we have ψ x = M and ψ = N. Then ψ = dx = x + C() = ψ = x + C () Since ψ = N = x e, we find C () = e. Using integration b parts, we find C() = e ( + ), so and our equation has solution Problem 18. Solve the following equation. ψ = x e ( + ) x e ( + ) = c. (3x + x + 3 ) dx + (x + ) d = 0. Solution. Computing the relevant partials, we find M = 3x + x + 3 N x = x. 1

13 Then M N x = 3 is a function of x alone (a constant, actuall), so we can compute an N integrating factor as follows. µ(x) = e 3 dx = e 3x. Multipling through b µ(x), we have M N {}}{{}}{ (3x e 3x + xe 3x + 3 e 3x ) dx + (x e 3x + e 3x ) d = 0 where we have redefined M and N. This equation is exact, so ψ = N for some function ψ(x, ). We determine ψ b integrating N with respect to : ψ = N d = (x e 3x + e 3x ) d = x e 3x e3x + C(x) for some function C(x). We also know ψ x = M, and taking the partial with respect to x of our expression for ψ, we find ψ x = xe 3x + 3x e 3x + 3 e 3x + C (x). Comparing this with M, we see C (x) = 0, so C(x) = C 0 is constant. Taking C 0 = 0, then we have the implicit solution x e 3x e3x = c. Problem 19. Find the general solution of the following differential equation. + 3 = 0 Solution. The characteristic equation is 0 = r + r 3 = (r + 3)(r 1), which has roots r = 3, 1. Then the general solution is = c 1 e 3t + c e t. Problem 0. (a) Solve the following initial value problem = 0, (0) = 4, (0) = 0 (b) Describe the solution s behavior as t. Solution. (a) The characteristic equation is 0 = 6r 5r + 1 = 6r r 3r + 1 = r(3r 1) (3r 1) = (r 1)(3r 1) which has roots r = 1/, 1/3. Then the general solution is The initial conditions impl = c 1 e t/ + c e t/3. 4 = (0) = c 1 + c 0 = (0) = c 1 + c 3. 13

14 Solving this sstem, we find c 1 = 8, c = 1, which ields the solution to the IVP = 8e t/ + 1e t/3. (b) Since the exponent on e t/ is larger, then as t. Problem 1. (a) Find the solution to the initial value problem 3 + = 0, (0) = (0) = 1/. (b) Determine the maximum value of the solution. Solution. (a) The characteristic equation is 0 = r 3r + 1 = r r r + 1 = r(r 1) (r 1) = (r 1)(r 1) which has roots 1/, 1. Then the general solution is The initial conditions impl = c 1 e t/ + c e t. = (0) = c 1 + c 1/ = (0) = c 1 / + c. Solving this sstem, we find c 1 = 3 and c = 1, so the solution to the IVP is (b) Setting = 3 et/ e t = 0, then = 3e t/ e t. 3 et/ = e t = 3 = et/ = ln(3/) = t/ = t = ln(3/). Since = 3 4 et/ e t and ( ln(3/)) = = = 9 8 < 0 this is a local maximum. Since this is the onl critical value, then it is the global maximum. Problem. Find the Wronskian of the given pairs of functions. (a) e t sin(t), (b) cos (θ), e t cos(t) 1 + cos(θ) Solution. (a) ( e W = t sin(t) e t ) cos(t) det e t (sin(t) + cos(t)) e t (cos(t) sin(t)) = e t (sin(t) cos(t) sin (t)) e t (cos(t) sin(t) + cos (t)) = e t (sin (t) + cos (t)) = e t 14

15 (b) B the double-angle formula, we have Problem cos(θ) = 1 + cos (θ) sin (θ) = cos (θ) since 1 sin (θ) = cos (θ). Then ( cos W = (θ) cos ) (θ) cos(θ) sin(θ) 4 cos(θ) sin(θ) = 4 cos 3 (θ) sin(θ) 4 cos 3 (θ) sin(θ) = 0 (a) Verif that 1 (t) = 1 and (t) = t are solutions to the differential equation + ( ) = 0 for t > 0. (b) Show that = c 1 + c t is not, in general, a solution to the above equation. (c) Explain wh this does not contradict Theorem 3... Solution. (a) The solution 1 (t) = 1 triviall satisfies the equation since it ields 0 = 0. We verif the second solution: + ( ) = t 1/ 1 ( ) 1 4 t 3/ + t 1/ = 1 4 t t 1 = 0. (b) Let = c 1 + c t 1/. Then so = c t 1/ = c 4 t 3/ + ( ) = (c 1 + c t 1/ ) c ( c 4 t 3/ + t 1/) = c 1c 4 t 3/ c 4 t 1 + c 4 t 1 = c 1c 4 t 3/ which is 0 if and onl if c 1 = 0 or c = 0. (c) Since the differential equation is nonlinear, Theorem 3.. does not appl. Problem 4. (a) Consider the differential equation + 4 = 0. Verif that the functions 1 = cos(t) and = sin(t) are solutions. Do the form a fundamental set of solutions? (b) Determine a fundamental set of solutions to the following differential equation = 0 15

16 Solution. (a) The characteristic equation is 0 = r + 4 which has roots r = ±i. Then we have solutions 1 (t) = e it = cos(t) + i sin(t) (t) = e it = cos(t) i sin(t). Taking real and imaginar parts, we have solutions u(t) = cos(t) and v(t) = sin(t). The Wronskian ( ) cos(t) sin(t) W = det = cos sin(t) cos(t) (t) + sin (t) = is not identicall 0 (in fact, it is never 0), so we have found a fundamental set of solutions. (b) The characteristic equation is 0 = r + 4r + 3 = (r + 3)(r + 1) which has roots r = 3, 1. Thus we have solutions 1 = e 3t, = e t. Since the Wronskian ( e 3t e W = t ) det = e 4t + 3e 4t = e 4t 3e 3t e t is never 0, these form a fundamental set of solutions. Problem 5. For each of the differential equations below, find the general solution. (a) + + = 0 (b) + 8 = 0 Solution. (a) The characteristic equation is 0 = r + r + which has solutions r = ± 4 8 = Thus the general solution is ± i = 1 ± i. = c 1 e t cos(t) + c e t sin(t). (b) The characteristic equation is 0 = r + r 8 = (r + 4)(r ) which has roots r = 4,. Thus the general solution is = c 1 e 4t + c e t. Problem 6. For each of the initial value problems below, find a solution and describe its behavior as t. (a) + 4 = 0, (0) = 0, (0) = 1 (b) + + (5/4) = 0, (0) = 3, (0) = 1 Solution. (a) The characteristic equation is 0 = r + 4 which has roots ±i. Thus the general solution is = c 1 cos(t) + c sin(t). We compute = c 1 sin(t) + c cos(t). 16

17 The initial conditions impl 0 = (0) = c 1 1 = (0) = c so c 1 = 0, c = 1/. Thus the solution to the IVP is = 1 sin(t). This solution is periodic with period π, and is bounded between 1/ and 1/. (b) The characteristic equation is 0 = r + r + 5/4 which has roots r = 1 ± 1 5 = Thus the general solution is 1 ± i = 1 ± i. = c 1 e t/ cos(t) + c e t/ sin(t). Note that 1 = c 1 e t/ cos(t) c 1 e t/ 1 sin(t) c e t/ sin(t) + c e t/ cos(t) (( = e t/ c ) ( 1 + c cos(t) + c 1 c ) ) sin(t). The initial conditions impl 3 = (0) = c 1 1 = (0) = c 1 + c so 1 = 3/ + c, hence c = 5/. Thus the solution to the IVP is = 3e t/ cos(t) + 5 e t/ sin(t). Because of the factors of e t/ in each term, the solution oscillates but with amplitude that decreases exponentiall and approaches 0 as t. 17