Problem 1 In each of the following problems find the general solution of the given differential

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1 VI Problem 1 dt + 2dy 3y = 0; dt 9dy + 9y = 0. Problem 2 dt + dy 2y = 0, y(0) = 1, y (0) = 1; dt 2 y = 0, y( 2) = 1, y ( 2) = Problem 3 Find the solution of the initial value problem 2 d2 y dt 2 3dy dt + y = 0, y(0) = 2, y (0) = 1 2. Then determine the maximum value of the solution and also find the point where the solution is zero. Problem 4 Solve the initial value problem dt 2 y = 0, y(0) = 2, y (0) = β. Then find β so that the solution approaches zero as t. 1

2 Problem 5 In the following (2α 1)dy + α(α 1)y = 0 dt determine the value of α, if any, for which all solutions tend to zero as t. Also determine the value of α, if any, for which all (nonzero) solutions become unbounded as t. Problem 6 where β > 0. 2 d2 y dt 2 + 3dy dt 2y = 0, y(0) = 1, y (0) = β Solve the initial value problem. Plot the solution when β = Find the coordinates (t 0, y 0 ) of the minimum point of the solution in this case. 3. Find the smallest value of β for which the solution has no minimum point. Problem 7 dt 2dy + 2y = 0; 9 dt 2 + 9dy dt 4y = 0. Problem 8 dt + 4y = 0, y(0) = 0, 2 y (0) = 1; dt 2 + 2dy dt + 2y = 0, y(π/4) = 2, y (π/4) = 2

3 Problem 9 3 d2 u dt 2 du dt + 2u = 0, u(0) = 2, u (0) = 0. Find the solution u(t) of this problem. Find the first time at which Problem 10 u(t) = 10. dt 2 + 2dy dt + 6y = 0, y(0) = 2, y (0) = α 0. Find the solution y(t) of this problem. Find α so that y = 0 when t = Problem 11 (Euler ) Solve the given for t > 0 Problem 12 t 2 d2 y dt 4tdy 6y = d2 y dt + 24dy + 9y = 0; dt 6dy + 9y = 0. Problem 13 dt 2 6dy dt + 9y = 0, y(0) = 0, y (0) = 2; dt 2 + 4dy dt + 4y = 0, y( 1) = 2, y ( 1) = 3

4 Problem 14 dt dy dt + 9y = 0, y(0) = 1, y (0) = 4 Solve the initial value problem and plot its solution for 0 t 5. Determine where the solution has the value zero. 3. Determine the coordinates (t 0, y 0 ) of the minimum point. 4. Change the second initial condition to y (0) = b > 0 and find the solution as a function of b. Then find those values of b that separate solutions that always remain positive from those that eventually become negative. Problem 15 dt 2 + 4dy dt + y = 0, y(0) = 1, y (0) = 2 Solve the initial value problem and plot the solution. Determine the coordinates (t 0, y 0 ) of the maximum point. 3. Change the second initial condition to y (0) = b > 0 and find the solution as a function of b. 4. Find the coordinates (t M, y M ) of the maximum point in terms of b. Describe the dependence of t M and y M on b as b increases. Problem 16 Use the method Abel formula together with the definition of the Wronskian to find a second solution of the given differential s (x 1) d2 y dx 2 xdy x 2 d2 y dx 2 + xdy dx + dx + y = 0, x > 1, y 1(x) = e x ; ( x ) y = 0, x > 0, y 1 (x) = sin x x ; 3. t d2 y dt 2 dy dt + 4t3 y = 0, t > 0, y 1 (t) = sin t 2. 4

5 Problem 17 (Behaviour of solutions as t ) If a, b, and c are positive constants show that all solutions of a d2 y dt + bdy + cy = 0 approach zero as t. (a) If a > 0 and c > 0 but b = 0 show that the result of the previous point is no longer true but that all solutions are bounded as t, that is if φ(t) is a solution verify that lim φ(t) K t where K is some positive constant. (b) If a > 0 and b > 0 but c = 0 show that the result of the point is no longer true, but that all solutions approach a constant that depends on the initial conditions as t. Determine this constant for the initial conditions y(0) = y 0, y (0) = y 0. (c) Show that y = sin t is a solution of dt + (k 2 sin2 t) dy + (1 k sin t cos t)y = 0 dt for any value of the constant k. If 0 < k < 2 show that 1 k cos t sin t > 0 and k sin 2 t 0. Thus observe that even though the coefficients of this variable coefficient differential are nonnegative (and the coefficient of y is zero only at the points t = 0, π, 2π, ) it has a solution that does not approach zero as t. Problem 18 (Turbulent Flows) The differential ( dx + δ x dy ) 2 dx + y = 0 arises in the study of the turbulent flow of a uniform stream past a circular cylinder. Verify that y 1 (x) = e δx2 /2 is one solution and then find the general solution in the form of an integral. 5