Page 1 of 5Student Name:Student Net ID:MATH 286 SECTION G1 – Introduction to Differential Equations PlusMIDTERM EXAMINATIONSeptember 20, 2012INSTRUCTOR: M. BRANNANINSTRUCTIONS• This exam 50 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.”Question: 1 2 3 4 5 TotalPoints: 12 8 10 10 10 50Score:Student Net ID: MATH 286 G1 Page 2 of 51. (a) (10 points) Solve the initial value problemxdydx− y = x cos2x + yx; y(e) = 0.(b) (2 points) Is the solution you found unique?Solution: (a). This is a homogeneous equation. Let v = y/x, then we have y =xv =⇒dydx= xdvdx+ v and thereforex(xdvdx+ v) = xv + x cos21 + v⇐⇒dvdx=cos2(1 + v)x⇐⇒ sec2(1 + v)dv =dxx=⇒ tan(1 + v) = ln |x| + C=⇒ v = arctan(ln |x| + C) − 1=⇒ y(x) = x arctan(ln |x| + C) − x.Using the IC y(e) = 0, we get0 = e arctan(1 + C) − e ⇐⇒ arctan(1 + C) = 1 ⇐⇒ C = tan(1) − 1.(b). The solution to the IVP we have found is unique by the existence-uniquenesstheorem. Indeed, we have y0= f(x, y), where f(x, y) =yx+ cos2x+yx. Since fand fyare obviously continuous on a rectangle containing initial condition (e, 0), thetheorem applies.2. (8 points) Find the general solution to the ODE1xdydx+ 2ex2y − ex2= 0 (x 6= 0).Solution: This equation is linear (it is also separable!): Multiplying by the inte-grating factor eP (x)= eR2xex2dx= eex2, this becomeseex2y0= xex2eex2⇐⇒ eex2y =12eex2+ C⇐⇒ y(x) =12+ Ce−ex2Student Net ID: MATH 286 G1 Page 3 of 53. (10 points) For what values of a and b is the equation(ye2xy+ x(1 + x) + y2)dx + (axy + bxe2xy)dy = 0exact? Find an implicit solution to this equation using these values of a and b.Solution: Let M = ye2xy+ x(1 + x) + y2and N = axy + bxe2xy. Then the givenequation is exact iffMy= e2xy+ 2xye2xy+ 2y = Nx= ay + be2xy+ 2bxye2xy, (x, y ∈ R).So a = 2 and b = 1.To find an implicit solution F (x, y(x)) = 0, we look for a function F (x, y) such thatFx= M and Fy= N. ThenF (x, y) =ZMdx =12e2xy+x33+x22+ xy2+ g(y),for some unknown function g(y). Differentiating F with respect to y, we getN = Fy= 2xy + xe2xy= xe2xy+ 2xy + g0(y),so g(y) = C is constant. Thus an implicit solution to this exact equation isF (x, y) = 0 ⇐⇒12e2xy+x33+x22+ xy2+ C = 0.4. (10 points) Find the general solution to the ODEdydt=ty + te−y− ye−t− e−y−t1 − e−yin implicit form.Solution: Factoring the right hand side we getdydt=(t − e−t)(y + e−y)1 − e−y,so this equation is separable. Separating variables, we have(1 − e−y)dyy + e−y= (t − e−t)dt=⇒ ln |y + e−y| =t22− e−t+ C⇐⇒ |y + e−y| = eCet22−e−tStudent Net ID: MATH 286 G1 Page 4 of 55. (10 points) A 1 kg rocket is launched from the ground at an angle of 0 < θ <π2radi-ans relative to the horizon, with an initial air speed of 10 m/s. Suppose that the dragforce on the rocket due to wind resistance is proportional to the rocket’s velocity. If theconstant of proportionality relating the drag to the velocity is k = 1 N· s/m and theacceleration due to gravity is −10 m/s2, what must the angle θ be so that the rocketreturns to the ground after 1 second of flight?(Note: Since you don’t have a calculator, an algebraic expression for θ is perfectlyfine.)Solution: Let v(t) denote the vertical component of the rocket’s velocity, so thatv0= v(0) = 10 sin θ m/s. Let h(t) denote the rocket’s height above the ground attime t, so that h0= h(0) = 0 and v(t) = h0(t) A vertical force balance then givesmdvdt= −mg − kv ⇐⇒dvdt= −10 − v.The integrating factor for this linear equation is eP (t)= eR1dt= et, sod(etv)dt= −10et=⇒ etv = −10et+ C ⇐⇒ v(t) = −10 + Ce−t.Solving for C using the initial condition, we getv(0) = 10 sin θ = −10 + C =⇒ C = 10 + 10 sin θ=⇒ v(t) = −10 + (10 + 10 sin θ)e−t.Since h(t) =Rv(t)dt = −10t − (10 + 10 sin θ)e−t+ K and h(0) = 0, we geth(t) = −10t + (10 + 10 sin θ)(1 − e−t).We now want to find the value of θ ∈ (0, π/2) such that h(1) = 0. Plugging in thesevalues, we obtain0 = −10 + (10 + 10 sin θ)(1 − e−1) ⇐⇒ θ = arcsin11 − e−1− 1= arcsin1e − 1.Student Net ID: MATH 286 G1 Page 5 of 5(Extra work