University of Connecticut Department of Mathematics University of Connecticut Department of Mathematics Math 2410 Exam 1, Spring 2014 Duration: 70 minutes Name: Section: Question Points 1 8 2 9 3 8 4 9 5 8 6 8 Total: 50 Score 1. Write clearly. Points may be deducted if your work is messy or your answer unclear; 2. You must show your work or explain your solution, otherwise points may be deducted; 3. Answer the questions in the space provided. You may use the back of the page if necessary; 4. No credit will be given for incorrect steps nor will credit be given for correct solutions arrived at by incorrect means; 1. (8 points) (a) Solve the initial value problem dy = t2 y 3 , y(0) = −1. dt (b) Solve the initial value problem dy = t2 y 3 , y(0) = 0. dt Note: you should give an explicit solution. Solution: (a) Z 1 dy = y3 Z t2 dt so −1 t3 = +c 2 2y 3 so, since y(0) = −1, we have c=− 1 2 and so −1 t3 1 = − 2y 2 3 2 or y2 = 3 3 − 2t3 so r y=− 3 . 3 − 2t3 (b) Note that the constant fucntion y(t) = 0 is also a solution to the given DE and is a solution to the given IVP. Page 2 2. (9 points) Solve the differential equation dy + 2y = sin 4t. dt using the method of undetermined coefficients (a.k.a. the method of ‘educated guessing’). Solution: The homogeneous equation dy + 2y = 0 dt has general solution yh = ce−2t . Suppose yp = α sin 4t + β cos 4t is a solution to the nonhomogeneous equation. Then dyp = 4α cos 4t − 4β sin 4t dt and so dyp + 2y = 4α cos 4t − 4β sin 4t + 2α sin 4t + 2β cos 4t dt thus we must have (2α − 3β) sin 4t + (4α + 2β) cos 4t = sin 4t and so α= so yp = 1 1 and β = − 10 5 1 1 sin 4t − cos 4t. 10 5 Thus, the general solution is y = yh + yp = ce−2t + Page 3 1 1 sin 4t − cos 4t. 10 5 3. (8 points) Solve the differential equation t dy + y = t2 + 1 dt using the method of integrating factors. Solution: When typing the exam, I accidentally put an easier equation here than I had intended to. Luck you! So, you may notice that the left-hand side here is already the derivative of a product t dy d +y = [ty] dt dt so, all we need to do is integrate Z ty = t2 + 1 dt = t3 +t+c 3 and solve for y y= t2 c +1+ . 3 t If you didn’t notice this, the usual method will still work. First, rewrite the equation into standard form dy 1 1 + y =t+ dt t t so, our integrating factor is R 1 e t dt = eln t = t and so dy + y = t2 + 1. dt We’re back to where we started, but now we realise that the left-hand side is the derivative of a product, so Z t3 ty = t2 + 1 dt = +t+c 3 or t2 c y= +1+ . 3 t t Page 4 4. (9 points) A cake is removed from an oven at 210◦ F and left to cool in a room which has a temperature of 70◦ F. After 10 minutes the temperature of the cake is 90◦ F. When will the temperature of the cake reach 80◦ F? Note: You may leave your answer in exact form. Solution: We know that the temperature T (t) is a solution to the differential equation dT = −k(T − 70) dt so T = ce−kt + 70 Now, since T (0) = 210, we have 140 = c so T = 140e−kt + 70 and since T (10) = 90, we have 20 = 140e−10k and so −kt e t 1 10 = 7 and so t 1 10 T = 140 + 70 7 Finally, to find when T (t) = 80: t 1 10 80 = 140 + 70 7 so 1 10 ln 14 . t= ln 71 Page 5 5. (8 points) Consider the autonomous differential equation dy = y(y − 2)4 (y + 1). dt (a) Find the equilibrium solutions. (b) Sketch the phase line. (c) Classify the equilibrium solutions as sources, sinks or nodes. Solution: (a) y = −1, y = 0 and y = 2 are the equilibrium solutions. y=2 y=0 y = −1 (b) (c) y = 2 is a node, y = 0 is a source and y = −1 is a sink. Page 6 6. (8 points) Consider the differential equation 1 dy = t2 y 5 . dt (a) Find the solution to the above equation. (b) Is there a unique solution to this differential equation such that y(0) = 0? (c) Is there a unique solution to this differential equation such that y(0) = 1? Note: for parts (b) and (c) you should justify your answer. Solution: (a) Z y − 15 Z dy = t2 dt so y=± 5 4 4 3 . t +c 15 Note also that y(t) = 0 is an equilibrium solution. (b) No. Note that 1 f (t, y) = t2 y 5 and ∂f t2 = 4 ∂y 4y 5 are discontinuous at (0, 0). Thus, the existence and unique theorem tells us nothing. Moreover, both y= 4 3 t 15 5 4 and y=0 are solutions of the IVP. (c) Yes. The functions f (t, y) and ∂f ∂y are both continuous at and around (0, 1), so the existence and uniqueness theorem guarantees a unique solution. Page 7