MT410 EXAM 1 SAMPLE 1 İLKER S. YÜCE DECEMBER 13, 2010 QUESTION 1. SOLUTIONS OF SOME DIFFERENTIAL EQUATIONS. dy dt = 4y 5, y(0) = y 0 (1) dy 4y 5 =

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1 MT EXAM SAMPLE İLKER S. YÜCE DECEMBER, SURNAME, NAME: QUESTION. SOLUTIONS OF SOME DIFFERENTIAL EQUATIONS where t. (A) Classify the given equation in (). = y, y() = y () (B) Solve the initial value problem. = y = y y = ln y = t + C, e ln y = e t+c y = Ce t y(t) = ±Ce t +. If y() = y, then y = ±Ce + ( or ±C = y ). ( The solution is y(t) = y ) e t +. (C) What can you say about the long-run behavior of y(t)? Note that the equilibrium solution is y =. Note also that if y > (, then y(t) = y ) e t + as t, if y < (, then y(t) = y ) e t + as t. (D) Determine which direction field belongs to the equation in (). By part (C), the answer is (a). (a) 6 8 (b) 6 8 (c) 6 8

2 QUESTION. LINEAR EQUATIONS; METHOD OF INTEGRATING FACTOR where t. (A) Classify the given equation in (). + y(t) = sin et, y() = y () (B) Solve the initial value problem. We have µ = e = e t t e + et y = e t sin e t d ( e t y ) = e t sin e t d ( e t y ) = e t sin e t e t y = u sin u du where u = e t. u sin u du = u( cos u) ( cos u) du u sin u du = u cos u + sin u + C. e t y(t) = e t cos e t + sin e t + C y(t) = e t cos e t + e t sin e t + Ce t. If y() = y, then y = Ce e cos + sin or C = y + cos sin. The solution is y(t) = (y + cos sin ) e t e t cos e t + e t sin e t. (C) Describe the behavior of the solutions as t. Note that we have lim y(t) = lim (y + cos sin ) e t e t cos e t + e t sin e t =. t t Therefore, the solutions y(t) are asymptotic to the t-axis. (The following is not required in the problem!) For example, if y =, we get

3 (A) Classify the equation in (). QUESTION. SEPARABLE EQUATIONS dx = x 8, y(9) =. () y (B) Solve the given initial value problem? (y ) = (x 8) dx (y ) = (x 8) dx y y = x 8x + C y y = x 6x + C y y + = x 6x + + C (y ) = x 6x + + C y = x 6x + + C y(x) = ± x 6x + + C. If y(9) = >, then = + 9 6(9) + + C or C = 6. The solution is y(x) = + (x 8) or y(t) = + x 8. (C) Determine the interval in which the solution is defined. Plot the graph of the solution. Note that y is not defined when y =. If y =, then we have = + x 8 x = 8. Therefore, the solution is not defined at (8, ). Then we have y(x) = x 6 if x > 8 or y(x) = x + if x < 8. Since the initial value is (9, ), the solution is y(x) = x 6 and the interval in which the solution is defined is (8, ). (9,)

4 QUESTION. SEPARABLE EQUATIONS: HOMOGENEOUS (A) Classify the differential equation in (). x dx y e y x + y =. () x First order, nonlinear, ordinary differential equation. (B) Show that the equation in () is homogeneous. dx = y e y x + x a (C) Solve the given differential equation. ( y x ) Let v = y/x. Then we get y = v + xv. In other words, we find that v + x dv dx = ( e v + v ). v The equation above is separable. We see that ve v dv = x dx ve v dv = x dx vev e v = ln x + C. We can obtain the solution only implicitly: y e x ( y ) x = ln x + C.

5 QUESTION. MODELING WITH FIRST ORDER EQUATIONS When a murder is committed, the bo, originally at 7 C, cools according to Newton s Law of Cooling. Suppose that after two hours the temperature is C, and the temperature of the surrounding air is a constant C. (Hint. Newton s Law of Cooling says that for some constant α, Rate of change of temperature = αtemperature difference.) (A) Find the temperature, H, of the bo as a function of t, the time in hours since the murder was committed. If H(t) is the temperature of the bo, then Temperature Difference=H(t). So, we derive dh = α(h(t) ). Since H() = 7 C, H(t) is decreasing. Therefore, α should be negative. Let us say α = k. Then we obtain the differential equation which models this event as: dh = k(h(t) ). The equation above is linear. By Integrating factor method, we find H(t) = Ce kt +. Since we have H() = 7, we get C = 7. We know that H() = or = 7e k or k.6. Thus, we conclude that H(t) = 7e.6t +. (B) Sketch a graph of temperature against time. H t (C) What happens to the temperature in the long run? Show this on the graph and algebraically. We calculate that lim H(t) = lim ( + t t 7e.6t ) =. Therefore, the temperature of the bo is going to approach the room temperature. (D) If the bo is found at pm at a temperature of C, when was the murder committed? We need to find t so that H(t ) =. We solve = + e.6t. We get t 8. hours = 8 hours minutes. Thus, the murder is committed at about 7:am.