Page 1 of 7Student Name:Student Net ID:MATH 286 SECTION X1 – Introduction to Differential Equations PlusMIDTERM EXAMINATION 1September 25, 2013INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam 60 minutes long. No personal aids or calculators are permitted.• Answer all questions in the space provided. If you require more space to write youranswer, you may continue on the back of the page. There is a blank page at the endof the exam for rough work.• EXPLAIN YOUR WORK! Little or no points will be given for a correct answer withno explanation of how you got it. If you use a theorem to answer a question, indicatewhich theorem you are using, and explain why the hypotheses of the theorem are valid.• GOOD LUCK!PLEASE NOTE: “Proctors are unable to respond to queries about the interpretation ofexam questions. Do your best to answer exam questions as written.”Student Net ID: MATH 286 X1 Page 2 of 7Question: 1 2 3 4 5 TotalPoints: 6 9 10 8 7 40Score:1. (6 points) Solve the initial value problemdydx= y−1eycos x, y(0) = 1.An implicit expression for the solution is fine.Solution: This is a separable equation:ye−ydy = cos xdx =⇒Zye−ydy =Zcos xdx + C =⇒ −ye−y− e−y= sin x + C.To find C, we plug in y(0) = 1, which gives −2e−1= C. The solution is thereforeye−y+ e−y= 2e−1− sin x.Student Net ID: MATH 286 X1 Page 3 of 72. Consider the ordinary differential equationdydx=yx+ x ln x (x > 0).(a) (3 points) Without explicitly solving this ODE, determine whether the correspond-ing initial value problem with initial condition y(1) = 0 has a unique solution, nosolution, or more than one solution. Explain your answer!Solution: Let F (x, y) =yx+ x ln x. According to the existence-uniquenesstheorem for initial value problems, there exists a unique solution to the aboveIVP if F and∂F∂y=1xare both continuous on a rectangle containing the initialdata (1, y(1)) = (1, 0). Since this is indeed the case, there is a unique solution.(b) (6 points) Find the solution to the IVPdydx=yx+ x ln x, y(1) = 1.Solution: We can write this equation asdydx−yx= x ln x, which is obviously linear,with integrating factoreP (x)= eR−x−1dx= e− ln x=1x.Multiplying through by this integrating factor, we get1xdydx−1x2y = ln x ⇐⇒ddxyx= ln x=⇒yx=Zln xdx + C = x ln x − x + C=⇒ y = x2ln x − x2+ Cx.Plugging in the initial conditions we get C = 2.Student Net ID: MATH 286 X1 Page 4 of 73. Consider the ODExx2+ y2+ 1− sin x +yx2+ y2+ 1dydx= 0 ((x, y) 6= (0, 0)).(a) (4 points) Show that this ODE is exact.Solution: Let M(x, y) =xx2+y2+1− sin x and N(x, y) =yx2+y2+1. ThenMy=−2xy(x2+ y2+ 1)2& Nx=−2xy(x2+ y2+ 1)2.Since Nx= My, the equation is exact.(b) (6 points) Find an implicit expression for the general solution to this ODE.Solution: We need to find a function F (x, y), defined for (x, y) 6= (0, 0), suchthat our solution curves lie along the level sets F (x, y) = C. If this is the case,then we must have M = Fxand N = Fy. Solving these equations, we obtainF (x, y) =ZM(x, y)dx =Zxx2+ y2+ 1− sin xdx=12ln(x2+ y2+ 1) + cos x + g(y),where g(y) is some unknown function of y. To find g(y), we use the equationFy= N, which givesN =yx2+ y2+ 1= Fy=yx2+ y2+ 1+ g0(y) =⇒ g0(y) = 0.Therefore g(y) = K is constant and our solutions are on the level curvesF (x, y) =12ln(x2+ y2+ 1) + cos x + K = C.Student Net ID: MATH 286 X1 Page 5 of 74. (8 points) Find the general solution to the second order ODExy00= x expy0x+ y0(x > 0),where exp a = ea. (Note: It is OK if your final answer involves an indefinite integralthat cannot be evaluated).Solution: Making the substitution z = y0, the equation becomes first order:xz0= x expzx+ z ⇐⇒ z0= expzx+zx.This is now a homogeneous substitution problem. Set v =zx. Then z0= xv0+ v,which givesxv0+ v = ev+ v =⇒ e−vdv =dxx=⇒ −e−v= ln x + C.Solving for v, we getzx= v = − ln(− ln x − C) =⇒ z = −x ln(− ln x − C)To get y, we integrate:y(x) =Zz(x)dx + K =Z−x ln(− ln x + C)dx + K.Student Net ID: MATH 286 X1 Page 6 of 75. (7 points) The functionsy1(x) = ex, y2(x) = e−x, y3(x) = e2x(x ∈ R),are solutions to the 3rd order linear ODEy000+ Ay00+ By0+ Cy = 0.What are A, B, C?Solution: The characteristic polynomial for this ODE is P (r) = r3+ Ar2+ Br + C.On the other hand, since y1, y2, y3are solutions, we know that P (r) has roots 1, −1,and 2 (each with multiplicity 1). ThusP (r) = (r − 1)(r + 1)(r − 2) = (r2− 1)(r − 2) = r3− 2r2− r + 2.Comparing coefficients, we obtainA = −2, B = −1, C = 2.Student Net ID: MATH 286 X1 Page 7 of 7(Extra work