Math 308 Exam I Practice Problems

Transcription

1 Math 308 Exam I Practice Problems This review should not be used as your sole source of preparation for the exam. You should also re-work all examples given in lecture and all suggested homework problems.. Find the general solution of the differential equation y + 3y = t + e 2t. Describe the behavior of solutions as t. 2. Solve the initial value problem ty = 2y + t 4 e t, y() = 0, t > 0. Describe how the solution behaves as t. 3. Find the general solution (in explicit form) of the differential equation y + y 2 sin x = Find the general solution (in implicit form) of the differential equation y = (cos 2 x)(cos 2 2y). 5. Find the solution of the initial value problem dx = y2 x, y() = 2 in explicit form and determine the interval in which the solution is defined. 6. A tank initially contains 50 lb of salt dissolved in 00 gal of water. Water containing 0.25 lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixture drains from the tank at the same rate. (a) Find an initial value problem for the amount of salt in the tank. (b) Solve the initial value problem in part (a) and find the amount of salt in the tank after a long time. (c) Find the time after which the salt level is within % of this limiting value. 7. Newton s Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that an object takes 40 minutes to cool from 30 C to 24 C in a room that is kept at 20 C. How long will it take the object to cool down to 2 C? 8. Suppose an object is launched straight up from the ground with initial velocity v 0. Ignoring the effects of air resistance, determine the time at which the object will return to the ground. (Your answer should be in terms of v 0 and the acceleration due to gravity at the earth s surface, g.)

2 9. A ball with mass 0. kg is thrown upward with initial velocity of 20 m/s from the roof of a building 30 m high. The force due to air resistance has a magnitude of v /30 directed opposite to the velocity. (a) Find an initial value problem for the velocity of the ball. (b) Solve the initial value problem in part (a) and determine the maximum height the ball reaches. 0. Determine (without solving the problem) an interval in which the solution of the initial value problem (t 2 3t)y + 2ty = ln t, y(4) = is certain to exist.. Consider the initial value problem dt = 3t2 + 4t + 2, y(0) =. 2(y ) State where in the ty-plane a unique solution is certain to exist. 2. Solve the initial value problem y + y 3 = 0, y(0) = y 0 and determine how the interval in which the solution exists depends on the value y Consider the differential equation y = y( y 2 ). Find all equilibrium solutions and determine their stability. 4. Consider the differential equation y = y 2 ( y) 2. Find all equilibrium solutions and determine their stability. 5. Find the general solution of the differential equation (y cos x + 2xe y ) + (sin x + x 2 e y )y = Find an integrating factor for the differential equation (x + 2) sin y + (x cos y)y = 0. Find the general solution of the differential equation. 2

3 7. Solve the initial value problem y + y 2y = 0, y(0) =, y (0) = 4 Describe the behavior of the solution as t. 8. Determine the longest interval in which the initial value problem (t 2 3t)y + ty (t + 3)y = 0, y() = 2, y () = is certain to have a unique twice-differentiable solution. 9. Verify that the functions y (t) = t and y 2 (t) = te t are solutions of the differential equation t 2 y t(t + 2)y + (t + 2)y = 0, t > 0. Do they form a fundamental set of solutions? 20. Find the Wronskian of two solutions of Legendre s equation without solving the equation. 2. Solve the initial value problem ( x 2 )y 2xy + α(α + )y = 0 y 4y + 3y = 0, y(0) =, y (0) = /3. Describe the behavior of the solution as t. 22. Solve the initial value problem y + 2y + y = 0, y(0) =, y (0) = 3. Describe the behavior of the solution as t. 23. Solve the initial value problem y + 2y + 5y = 0, y(0) = 2, y (0) =. Describe the behavior of the solution as t. 24. Verify that y (t) = e t is a solution of (t )y ty + y = 0, t >. Use the method of reduction of order to find a fundamental set of solutions. 3

4 Solutions Solutions may contain errors or typos. If you find an error or typo, please notify me at Find the general solution of the differential equation y + 3y = t + e 2t. Describe the behavior of solutions as t. To solve, we first compute the integrating factor ( ) µ(t) = exp 3 dt = e 3t. Multiplying by µ(t), we obtain e 3t y + 3e 3t y = te 3t + e t d dt (e3t y) = te 3t + e t. Integrating both sides of this equation, we have e 3t y = (te 3t + e t ) dt Therefore, the general solution is As t, solutions diverge to. e 3t y = 3 te3t 9 e3t + e t + C. y(t) = 3 t 9 + e 2t + Ce 3t. 4

5 2. Solve the initial value problem ty = 2y + t 4 e t, y() = 0, t > 0. Describe how the solution behaves as t. To solve, we first express the equation in standard form ty 2y = t 4 e ( ) t 2 y y = t 3 e t. t Then we find an integrating factor ( µ(t) = exp 2 t dt ) = e 2 ln t = t 2. Multiplying by µ(t), we obtain t 2 y 2t 3 y = te t d ( t 2 y ) dt = te t. Integrating both sides of this equation, we have t 2 y = te t dt Therefore, the general solution is t 2 y = te t e t + C t 2 y = (t )e t + C. y(t) = (t 3 t 2 )e t + Ct 2. Applying the initial condition, we obtain y() = C = 0. Thus, the solution is As t, the solution diverges to. y(t) = (t 3 t 2 )e t + 0t 2. 5

6 3. Find the general solution (in explicit form) of the differential equation Using separation of variables, we obtain y + y 2 sin x = 0. dx = y2 sin x = sin x dx (y 0) y 2 y = sin x dx 2 = cos x + C. y Therefore, the general solution (in explicit form) is Note that y(x) = 0 is a solution. y(x) = C cos x. 6

7 4. Find the general solution (in implicit form) of the differential equation Using separation of variables, we obtain If cos(2y) = 0, then y = (cos 2 x)(cos 2 (2y)). dx = (cos2 x)(cos 2 (2y)) cos 2 (2y) = (cos 2 x) dx (cos(2y) 0) (sec 2 (2y)) = (cos 2 x) dx sec 2 (2y) = cos 2 x dx sec 2 (2y) = 2 ( + cos(2x)) dx 2 tan(2y) = x sin(2x) + C 2 tan(2y) x 2 4 sin(2x) = C 2 tan(2y) 2x sin(2x) = C. 2y = y = (2n + )π 2 (2n + )π 4 Note that y(x) = (2n + )π/4 is a solution for any integer n. 7

8 5. Find the solution of the initial value problem dt = y2 t, y() = 2 in explicit form and determine the interval in which the solution is defined. Using separation of variables, we obtain = dt y 2 t y = ln t + C y(t) = C ln t. Applying the initial condition, we obtain y() = C = 2. It follows that C = /2, and so y(t) = The solution is defined if t > 0 and /2 ln t = 2 2 ln t. 0 < 2 ln t 2 ln t < ln t < 2 t < e /2. Thus, the solution is defined in the interval 0 < t < e. 8

9 6. A tank initially contains 50 lb of salt dissolved in 00 gal of water. Water containing 0.25 lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixture drains from the tank at the same rate. Find the amount of salt in the tank after a long time. Find the time after which the salt level is within % of this limiting value. (a) Find an initial value problem for the amount of salt in the tank. Let y(t) denote the amount (in lb) of salt in the tank after t 0 minutes. Thus, dt dt dt = (0.25 lb/gal)(3 gal/min) ( = y ) lb/min 00 = 3 (25 y). 00 Therefore, we have the initial value problem dt = 3 (25 y), y(0) = ( y 00 lb/gal ) (3 gal/min) (b) Solve the initial value problem in part (a) and find the amount of salt in the tank after a long time. Using separation of variables, we obtain = 3 25 y 00 dt 25 y = 3 00 dt ln 25 y = 0.03t + C 25 y = Ce 0.03t Applying the initial condition, we obtain y(t) = 25 Ce 0.03t. y(0) = 25 C = 50. Thus, C = 25 and the solution of the initial value problem is y(t) = 25( + e 0.03t ). As t, the the amount of salt in the tank approaches 25 lb. 9

10 (c) Find the time after which the salt level is within % of this limiting value. To find the time at which the salt level is within % of this limiting value, let 25( + e 0.03t ) = (.0)(25) + e 0.03t =.0 e 0.03t = t = ln(0.0) t = ln(0.0) 0.03 t 53.5 min t 2 hr 36 min. Note: On the exam, you will not need to convert the unit of time. 0

11 7. Newton s Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that an object takes 40 minutes to cool from 30 C to 24 C in a room that is kept at 20 C. How long will it take the object to cool down to 2 C? Let y(t) denote the temperature (in C) after t 0 minutes. By Newton s Law of Cooling, = k(20 y), y(0) = 30. dt Using separation of variables, we obtain = k dt 20 y 20 y = k dt ln 20 y = kt + C Applying the initial condition, we obtain 20 y = Ce kt y(t) = 20 Ce kt. y(0) = 20 C = 30. Thus, C = 0 and the solution of the initial value problem is We are given that y(40) = 24, so y(t) = e kt. Therefore, e 40k = 24 0e 40k = 4 e 40k = k = ln(0.4) k = ln(0.4) 40. y(t) = e ln(0.4)t/40.

12 To find the time at which the object reaches a temperature of 2 C, let e ln(0.4)t/40 = 2 0e ln(0.4)t/40 = e ln(0.4)t/40 = 0. ln(0.4) t 40 = ln(0.) t = 40 ln(0.) ln(0.4) t 00.5 min t h 4 min. 2

13 8. Suppose an object is launched straight up from the ground with initial velocity v 0. Ignoring the effects of air resistance, determine the time at which the object will return to the ground. (Your answer should be in terms of v 0 and the acceleration due to gravity at the earth s surface, g.) Let v(t) denote the velocity (in m/s) of the object at time t 0, and assume that a positive velocity corresponds to movement upward. By Newton s Second Law, F = ma = m dv dt. Ignoring air resistance, the only force acting on the object is the force due to gravity. That is, F = mg. Therefore, m dv dt dv dt = mg = g. Integrating with respect to t, the velocity of the object at time t is given by v(t) = gt + v 0. Integrating with respect to t again, the height of the object above the ground (in m) at time t is given by h(t) = g 2 t2 + v 0 t. To determine the time at which the object will return to the ground, let g 2 t2 + v 0 t = 0 t (v 0 g ) 2 t = 0 t = 2v 0 g s Note: We take the positive value of t since t = 0 corresponds to the time at which the ball was initially launced upward. 3

14 9. A ball with mass 0. kg is thrown upward with initial velocity of 20 m/s from the roof of a building 30 m high. The force due to air resistance has a magnitude of v /30 directed opposite to the velocity. (a) Find an initial value problem for the velocity of the ball. Let v(t) denote the velocity (in m/s) of the ball after t 0 seconds, and assume that a positive velocity corresponds to the upward direction. By Newton s Second Law, F = ma = m dv dt. The force due to gravity acting on the ball is F g = mg and the force due to air resistance is F a = v/30. Therefore, the net force acting on the ball is Therefore, Newton s Second Law yields F = F g + F a = mg v 30. m dv dt (0.) dv dt dv dt = mg v 30 = (0.)(9.8) v 30 = 9.8 v 3 dv dt + v 3 = 9.8. Therefore, an initial value problem for the velocity of the ball is dv dt + v 3 = 9.8, v(0) = 20. 4

15 (b) Solve the initial value problem in part (a) and determine the maximum height the ball reaches. To solve the initial value problem, we first find an integrating factor Multiplying by µ(t), we obtain Therefore, the velocity of the ball is µ(t) = e /3 dt = e t/3. e t/3 v + 3 et/3 v = 9.8e t/3 d ( e t/3 v ) = 9.8e t/3 dt e t/3 v = 9.8 e t/3 dt e t/3 v = 29.4e t/3 + C. v(t) = Ce t/3. Applying the initial condition, we obtain v(0) = C = 20. It follows that C = 49.4 and so the velocity of the ball is v(t) = e t/3 m/s. To determine the height of the ball, h(t), we integrate the velocity function h(t) = e t/3 dt h(t) = 29.4t 48.2e t/3 + C. Applying the initial condition, we obtain h(0) = C = 30. It follows that C = 78.2 and the height of the ball is h(t) = 29.4t 48.2e t/ m. 5

16 To determine the time at which the ball reaches its maximum height, we set the velocity function equal to zero. That is, e t/3 = e t/3 = 29.4 e t/3 = e t/3 = ( ) t = ln 29.4 ( ) 494 t = 3 ln s Substituting t =.6 into the height function, the maximum height of the ball is h(.6) m. Note: On the exam, you will not be expected to evaluate messy expressions such as h(.6). 0. Determine (without solving the problem) an interval in which the solution of the initial value problem (t 2 3t)y + 2ty = ln t, y(4) = is certain to exist. In standard form, we have y + 2t t 2 3t y = ln t t 2 3t. So p(t) = 2t/(t 2 3t) and g(t) = ln t/(t 2 3t). Thus, for this equation, p is continuous for t 0, 3 and g is continuous for t > 0 and t 3. Both p and g are continuous on (0, 3) (3, ). The interval containing the initial point t 0 = 4 is (3, ). By Theorem 2.4., the initial value problem has a unique solution on the interval (3, ). 6

17 . Consider the initial value problem dt = 3t2 + 4t + 2, y(0) =. 2(y ) State where in the ty-plane a unique solution is certain to exist. For this equation, f(t, y) = 3t2 + 4t + 2, 2(y ) f y (t, y) = + 4t + 2 3t2 2(y ). 2 Each of these functions is continuous for y. That is, f and f/ y are continuous on the set {(t, y) y }. Since y(0) =, we consider the region in the ty-plane defined by {(t, y) y < }. By Theorem 2.4.2, the initial value problem has a unique solution within the set {(t, y) y < }. 7

18 2. Solve the initial value problem y + y 3 = 0, y(0) = y 0 and determine how the interval in which the solution exists depends on the value y 0. Using separation of variables, we obtain dt y = 3 Applying the initial condition, we obtain It follows that C = /y 2 0 and so = y 3 = dt y 3 dt 2y 2 = t + C y 2 = 2t + C y 2 = y 2 0 = C. 2t + C. y 2 = y 2 = y = 2t + /y 2 0 y 2 0 2y0t 2 + y 0 2y 2 0 t +. This solution is only valid if y 0 0. Therefore, if y 0 0, the solution exists if and only if 2y0t 2 + > 0 t >. 2y0 2 If y 0 = 0, then y = 0 and y(t) = 0 for all t R. 8

19 3. Consider the differential equation y = y( y 2 ). Find all equilibrium solutions and determine their stability. To find the equilibrium solutions, let y = y( y 2 ) = 0. Thus, the equilibrium solutions are ȳ =, 0,. To determine their stability, sketch the graph of y versus y. If y <, then solutions are increasing. If < y < 0, then solutions are decreasing. If 0 < y <, then solutions are increasing. If y >, then solutions are decreasing. Therefore, ȳ = is asymptotically stable, ȳ = 0 is unstable, and ȳ = is asymptotically stable. 9

20 4. Consider the differential equation y = y 2 ( y) 2. Find all equilibrium solutions and determine their stability. To find the equilibrium solutions, let y = y 2 ( y) 2 = 0. Thus, the equilibrium solutions are ȳ = 0,. To determine their stability, sketch the graph of y versus y. If y < 0, then solutions are increasing. If 0 < y <, then solutions are increasing. If y >, then solutions are increasing. Therefore, ȳ = 0 and ȳ = are both semistable. 20

21 5. Find the general solution of the differential equation (y cos x + 2xe y ) + (sin x + x 2 e y )y = 0. Let M(x, y) = y cos x + 2xe y and N(x, y) = sin x + x 2 e y. Thus, M y (x, y) = cos x + 2xe y = N x (x, y). Therefore, the equation is exact and there exists a function ψ(x, y) such that ψ x (x, y) = y cos x + 2xe y ψ y (x, y) = sin x + x 2 e y. Integrating the first equation with respect to x, we obtain ψ(x, y) = (y cos x + 2xe y ) dx = y sin x + x 2 e y + f(y). Differentiating with respect to y, we obtain ψ y (x, y) = sin x + x 2 e y + f (y). Therefore, sin x + x 2 e y + f (y) = sin x + x 2 e y f (y) = f(y) = y. Then ψ(x, y) = y sin x + x 2 e y y and solutions are defined implicitly by y sin x + x 2 e y y = C. 2

22 6. Find an integrating factor for the differential equation (x + 2) sin y + (x cos y)y = 0. Find the general solution of the differential equation. Upon calculating (M y N x )/N, we obtain M y N x N = (x + 2) cos y cos y x cos y = (x + ) cos y x cos y = x + x = + x. Thus, there is an integrating factor given by [ ( µ(x) = exp + ) ] dx x = e x+ln x = xe x. Multiplying the equation by this integrating factor, we obtain (x 2 + 2x)e x sin y + (x 2 e x cos y)y = 0. Since this equation is exact, there exists a function ψ(x, y) such that ψ x (x, y) = (x 2 + 2x)e x sin y ψ y (x, y) = (x 2 e x cos y). Integrating the second equation with respect to y, we have ψ(x, y) = (x 2 e x cos y) = x 2 e x sin y + f(x). Differentiating with respect to x, we obtain ψ x (x, y) = (2x + x 2 )e x sin y + f (x). Therefore, (2x + x 2 )e x sin y + f (x) = (x 2 + 2x)e x sin y f (x) = 0 f(x) = C = 0. Then ψ(x, y) = x 2 e x sin y and solutions are defined implicitly by x 2 e x sin y = C. 22

23 7. Solve the initial value problem y + y 2y = 0, y(0) =, y (0) = 4 Describe the behavior of the solution as t. Let y = e rt. Then r must be a root of the characteristic equation r 2 + r 2 = (r + 2)(r ) = 0 The roots of the characteristic equation are r = 2 and r 2 =. Thus, the general solution (and its derivative) are y(t) = c e 2t + c 2 e t, Applying the initial conditions, we obtain y (t) = 2c e 2t + c 2 e t. y(0) = c + c 2 =, y (0) = 2c + c 2 = 4. It follows that c = and c 2 = 2. Therefore, the solution is As t, the solution diverges to. y(t) = e 2t + 2e t. 8. Determine the longest interval in which the initial value problem (t 2 3t)y + ty (t + 3)y = 0, y() = 2, y () = is certain to have a unique twice-differentiable solution. In standard form, we have y + ( ) ( ) t + 3 y y = 0. t 3 t 2 3t So p(t) = /(t 3), q(t) = (t + 3)/(t 2 3t), and g(t) = 0 are both continuous for t 0, 3. Therefore, the longest open interval containing t 0 = in which the coefficients are continuous is (0, 3). By Theorem 3.2., a unique solution exists in (0, 3). 23

24 9. Verify that the functions y (t) = t and y 2 (t) = te t are solutions of the differential equation t 2 y t(t + 2)y + (t + 2)y = 0, t > 0. Do they form a fundamental set of solutions? If y (t) = t, then y (t) = and y (t) = 0. So t 2 y t(t + 2)y + (t + 2)y = t 2 (0) t(t + 2)() + (t + 2)(t) = 0. If y 2 (t) = te t, then y 2(t) = ( + t)e t and y 2(t) = (2 + t)e t. So t 2 y 2 t(t + 2)y 2 + (t + 2)y 2 = t 2 (2 + t)e t t(t + 2)( + t)e t + (t + 2)te t = 0. Thus, y and y 2 are solutions of the differential equation. The Wronskian of y and y 2 is W (y, y 2 )(t) = t tet ( + t)e t = (t + t2 )e t te t = t 2 e t. Since W 0 for t > 0, y and y 2 form a fundamental set of solutions. 20. Find the Wronskian of two solutions of Legendre s equation without solving the equation. ( x 2 )y 2xy + α(α + )y = 0 In standard form, we have y ( ) 2x y + x 2 ( α(α + ) x 2 ) y = 0. By Abel s Theorem, the Wronskian of two solutions is [ ] 2x W (y, y 2 )(x) = c exp x dx = ce ln( x2) = c 2 x. 2 Note: To evaluate the integral above, we use a substitution of u = x 2. 24

25 2. Solve the initial value problem y 4y + 3y = 0, y(0) =, y (0) = /3. Describe the behavior of the solution as t. Let y = e rt. Then r must be a root of the characteristic equation r 2 4r + 3 = (r )(r 3) = 0. The roots of the characteristic equation are r = and r = 3. Therefore, the general solution (and its derivative) are Applying the initial conditions, we obtain y(t) = c e t + c 2 e 3t y (t) = c e t + 3c 2 e 3t y(0) = c + c 2 = y (0) = c + 3c 2 = /3 It follows that c = 4/3 and c 2 = /3. Thus, the solution of the initial value problem is y(t) = 4 3 et 3 e3t. As t, the solution diverges to y t 25

26 22. Solve the initial value problem y + 2y + y = 0, y(0) =, y (0) = 3. Describe the behavior of the solution as t. Let y = e rt. Then r must be a root of the characteristic equation r 2 + 2r + = (r + ) 2 = 0. There is a single, repeated root r = r 2 =. Therefore, the general solution (and its derivative) are y(t) = c e t + c 2 te t Applying the initial conditions, we obtain y (t) = c e t + ( t)c 2 e t y(0) = c = y (0) = c + c 2 = 3 It follows that c = and c 2 = 2. Thus, the solution of the initial value problem is As t, the solution converges to 0. y(t) = e t 2te t = ( 2t)e t y t 26

27 23. Solve the initial value problem y + 2y + 5y = 0, y(0) = 2, y (0) =. Describe the behavior of the solution as t. Let y = e rt. Then r must be a root of the characteristic equation r 2 + 2r + 5 = 0. The roots of the characteristic equation are r = 2 ± = ± 2i Therefore, the general solution is y(t) = c e t cos(2t) + c 2 e t sin(2t) Differentiating, we obtain y (t) = c e t cos(2t) 2c e t sin(2t) c 2 e t sin(2t) + 2c 2 e t cos(2t) Applying the initial conditions, we obtain y(0) = c = 2 y (0) = c + 2c 2 = It follows that c = 2 and c 2 = 3/2. Thus, the solution of the initial value problem is y(t) = 2e t cos(2t) e t sin(2t). As t, the solution exhibits a decaying oscillation approaching zero y t 27

28 24. Verify that y (t) = e t is a solution of (t )y ty + y = 0, t >. Use the method of reduction of order to find a fundamental set of solutions. If y (t) = e t, then y (t) = y (t) = e t. Then (t )y ty + y = (t )e t te t + e t = 0. Thus, y (t) = e t is a solution of the differential equation. Setting y(t) = v(t)e t, then y (t) = v e t + ve t, y (t) = v e t + 2v e t + ve t. Substituting y, y, and y in the differential equation and collecting terms, we obtain (t )y ty + y = (t )(v e t + 2v e t + ve t ) t(v e t + ve t ) + ve t = (t )e t v + (t 2)e t v + ((t )e t te t + e t )v = (t )e t v + (t 2)e t v = 0. Note that the coefficient of v is zero, as it should be. Setting w(t) = v (t), then Using separation of variables, we obtain Then It follows that (t )e t w + (t 2)e t w = 0. dw w = 2 t t dt ( ) w dw = t dt ln w = ln t t + C v(t) = w(t) = C(t )e t. C(t )e t dt = Cte t + K. y(t) = v(t)e t = Ct + Ke t, where C and K are arbitrary constants. The second term is a multiple of y (t) = e t and can be dropped, but the first term produces a new solution y 2 (t) = t. The Wronskian of y and y 2 is W (y, y 2 )(t) = et t e t = ( t)et 0 for t >. Therefore, y (t) = e t and y 2 (t) = t form a fundamental set of solutions of the differential equation for t >. 28