Math 308 Exam I Practice Problems

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1 Math 308 Exam I Practice Problems This review should not be used as your sole source for 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 and use it to determine how solutions behave as t. 2. Solve the initial value problem y t =, y() =. t + 3. Find the general solution of the differential equation y + y 2 sin x = Find the general solution 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. 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. 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.) 9. 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.

2 0. Consider the initial value problem = 3t2 + 4t + 2, y(0) =. 2(y ) State where in the ty-plane a unique solution is certain to exist.. 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. 3. Consider the differential equation y = y 2 ( y) 2. Find all equilibrium solutions and determine their stability. 4. Consider the single-species population model dn with initial condition N(0) = N 0 0. = 2N(N 50) ( N ), 400 (a) Find all equilibrium solutions of this differential equation and determine their stability. (b) If N 0 = 00, find lim N(t). t (c) If N 0 = 200, find lim N(t). t (d) If N 0 = 500, find lim N(t). t 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 and describe how the solution behaves 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. ( x 2 )y 2xy + α(α + )y = 0 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 and use it to determine how solutions behave as t. To solve, we first compute the integrating factor ( ) µ(t) = exp 3 = e 3t. Multiplying by µ(t), we obtain e 3t y + 3e 3t y = te 3t + e t d (e3t y) = te 3t + e t. Integrating both sides of this equation, we have e 3t y = (te 3t + e t ) 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 y t =, y() =. t + To solve, we first compute the integrating factor ( µ(t) = exp ) t = e ln t = t. Multiplying by µ(t), we obtain t y t y = 2 t(t + ) d ( y ) = t t(t + ). Integrating both sides of this equation, we have y = t t(t + ) ( y = t t ) t + y = ln(t) ln(t + ) + C t ( ) y t = ln + C t t + Therefore, the general solution is y(t) = t ln Applying the initial condition, we obtain It follows that C = + ln 2, and so y(t) = t ln ( ) t + Ct. t + y() = ln 2 + C =. ( ) t + ( + ln 2)t. t + 5

6 3. Find the general solution of the differential equation Using separation of variables, we obtain y + y 2 sin x = 0. dx = y2 sin x = sin x dx y 2 y = sin x dx 2 = cos x + C. y Therefore, the general solution (in explicit form) is y(x) = C cos x. 4. Find the general solution of the differential equation Using separation of variables, we obtain y = (cos 2 x)(cos 2 (2y)). dx = (cos2 x)(cos 2 (2y)) cos 2 (2y) = (cos 2 x) dx (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. The solution is defined implicitly. 6

7 5. Find the solution of the initial value problem = y2 t, y() = 2 in explicit form and determine the interval in which the solution is defined. Using separation of variables, we obtain = 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. 7

8 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. Let y(t) denote the amount (in lb) of salt in the tank at time t 0. Thus, ( y ) = (0.25 lb/gal)(3 gal/min) 00 lb/gal (3 gal/min) ( = y ) lb/min 00 = 3 (25 y). 00 Therefore, we have the initial value problem = 3 (25 y), y(0) = Using separation of variables, we obtain = 3 25 y y = 3 00 ln 25 y = 0.03t + C Applying the initial condition, we obtain 25 y = Ce 0.03t 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 a limiting value of 25 lb. 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. 8

9 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) at time t 0. By Newton s Law of Cooling, = k(20 y), y(0) = 30. Using separation of variables, we obtain = k 20 y 20 y = k 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 Therefore, y(t) = e kt 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. 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 9 t h 4 min.

10 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. Ignoring air resistance, the only force acting on the object is the force due to gravity. That is, F = mg. Therefore, m dv dv = 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. 0

11 9. 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. 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, ). 0. Consider the initial value problem = 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 everywhere except on the line y =. Thus, a rectangle can be drawn around the initial point (0, ) in which both f and f y are continuous. By Theorem 2.4.2, there is a unique solution in some interval about the initial point t 0 = 0.

12 . 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 y = 3 Applying the initial condition, we obtain It follows that C = /y 2 0 and so = y 3 = y 3 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. 2

13 2. 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. 3

14 3. 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. 4

15 4. Consider the single-species population model dn with initial condition N(0) = N 0 0. = 2N(N 50) ( N ), 400 (a) Find all equilibrium solutions of this differential equation and determine their stability. To find the equilibrium solutions, let dn = 2N(N 50) ( N ) = Thus, the equilibrium solutions are N = 0, 50, 400. To determine their stability, sketch the graph of dn/ versus N. If 0 < N < 50, then solutions are decreasing. If 50 < N < 400, then solutions are increasing. If N > 400, then solutions are decreasing. Therefore, N = 0 is asymptotically stable, ȳ = 50 is unstable, and ȳ = 400 is asymptotically stable. (b) If N 0 = 00, find lim t N(t). If N 0 = 00, then N(t) 0. (c) If N 0 = 200, find lim t N(t). If N 0 = 200, then N(t) 400. (d) If N 0 = 500, find lim t N(t). If N 0 = 500, then N(t)

16 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. 6

17 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. 7

18 7. Solve the initial value problem y + y 2y = 0, y(0) =, y (0) = 4 and describe how the solution behaves as t. The characteristic equation is r 2 + r 2 = 0 (r + 2)(r ) = 0 r = 2,. Thus, the general solution (and its derivative) are y(t) = c e 2t + c 2 e t, y (t) = 2c e 2t + c 2 e t. Applying the initial conditions, we obtain y(0) = c + c 2 =, y (0) = 2c + c 2 = 4. It follows that c = and c 2 = 2. Therefore, the solution is y(t) = e 2t + 2e t. As t, the solution diverges to. 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 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). 8

19 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. 9

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