N e t I D : N a m e :Midterm#1MATH 286—Differential Equations PlusThursday, February 13• No notes, p ersonal aids or calculator s are permitted.• Answer all questions in the space provided. If you require more space to write your answer, you may continueon the back of the page. There is a blank page at the end of the exam for rough work.• Explain your work! Little or no points will be given for a correct answer with no explanation of h ow you got it.Good luck!Problem 1. (5 points) Circle the slope field below which belongs to the differ e ntial equation exy′= x −y.-3-2-10123-3-2-10123-3-2-10123-3-2-10123-3-2-10123-3-2-10123Problem 2. (20 points) F ind the general solution to the differential equation y(5)−4y(4)+ 4y(3)= 0.Solution. This is a linear homogeneous equation with constant coefficients. The characteristic polynomial is r5−4r4+ 4r3= r3(r −2)2.Hence, the general solution is c1+ c2x + c3x2+ (c4+ c5x)e2x. Michael Brannan Armin [email protected] [email protected]/4Problem 3. (20 points) Solve the initial value problem(x2+ 1)dydx+ xy =1x2+ 1√, y(0) = 1.Solution. Dividing by x2+ 1, we getdydx+xx2+ 1y =1(x2+ 1)3/2,which is a linear equation with integrating factoreRxx2+1dx= e12ln (x2+1)= (x2+ 1)1/2.Multiply through by this integrating factor, we obtain(x2+ 1)1/2dydx+x(x2+ 1)1/2y =ddx(x2+ 1)1/2y=1x2+ 1.Integrating both sides wit h respect to x yields(x2+ 1)1/2y(x) = arctan (x) + C.Since y(0) = 1, we find C = 1. Therefore,y(x) =arctan (x) + 1(x2+ 1)1/2. Problem 4. (20 points) The time rate of change of a rabbit population P is proportional to the square root of P .At time t = 0, the population numbers 100 rabbits and is increasing at the rate of 20 rabbits per month. How manyrabbits will there be after two months?Solution.dPdt= k P√and P (0) = 100, P′(0) = 20. The problem asks for P (2).At t = 0, we have 20 = P′(0) = k P (0)p= 10k. Hence k = 2.By separation of variables, we find 2 P√= 2t + C. From P (0) = 100, C = 20. Thus, P = (t + 10)2.We conclude that there are P (2) = 122= 144 rabbits after two months. Michael Brannan Armin [email protected] [email protected]/4Problem 5. (20 points) For each c > 0, let yc(x) =x3, if x < 0,0, if 0 6 x 6 c,(x −c)3, if x > c.(a) Sketch the graph of yc(x) for c = 3/2.(b) Show that, for all c > 0, ycis a solution to the initial value problemdydx= 3y2/3, y(0) = 0.(c) Explain why (b) does not contradict the theorem on e xistence and uniqueness for solutions to init ial valueproblems.Solution.(a) For c = 3/2, the graph looks as follows.-1123c-2246(b) Clearly, yc(0) = 0. Moreover, sinceddxx3= 3x2= 3(x3)2/3,ddx(x −c)3= 3(x −c)2= 3[(x −c)3]2/3, and all the leftand right derivatives at the appropriate endpoints ma tch, yc(x) satisfies the ODE y′= 3 y2/3.(c) Write the DE as y′= f(x, y) with f (x, y) = 3y2/3. Since∂∂yf(x, y) = 2y−1/3is not continuous at the point(0, 0) = (0, y(0)), the hypotheses of the theorem are not satisfied for this initial value problem. Michael Brannan Armin [email protected] [email protected]/4Problem 6. (20 points) F ind a general solution to the differential equa tionx2dydx−x2− y2−3xy = 0.Solution. Dividing through by x2, the equation becomesdydx= 1 + (y/x)2+ 3 (y/x).Making the homogeneous substitution v = y/x, we havedydx= xdvdx+ v = 1 + v2+ 3v, which givesxdvdx= v2+ 2 v + 1 = (v + 1)2.By separation of variables, we getdv(v + 1)2=dxx(note that we just lost the singular solution v = −1 which correspondsto the solution y = −x of the original equation). Integrating, we have −(v + 1)−1= ln |x|+ C and hencey(x) = −xln |x|+ C−x. Problem 7. (5 points) Consider the differential equation Hint: Do not attempt to solve the DE.y′= y4+ x4+ 1.Is it possible that there exists a solution with th e property that limx→ ∞y(x) = −∞? Why, or why not?Solution. It is im possible. It follows from the differential equation t hat y′(x) > 0 for all x. I n other words, any solutiony(x) is an increasing function, and thus cannot approach −∞ for large x. Michael Brannan Armin [email protected]