N e t I D : N a m e :Midterm#2MATH 286—Differential Equations PlusThursday, March 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) Consider a homogeneous linear differential equation with consta nt real coefficients whichhas order 6. Suppose y(x) = x2e2xcos(x) is a solution. Write down the general solution.Solution. The general solution is (c1+ c2x + c3x2)e2xcos(x) + (c4+ c5x + c6x2)e2xsin(x). Problem 2. (20 points) Find the general solution of x′=1 22 1x.Solution. The characteristic polynomial (1 −λ)2−4 has roots λ = −1, 3.For λ = 3, solving−2 22 −2x =00, we find the eigenve ctor v =11.For λ = −1, solving2 22 2x =00, we find the eigenvector v =1−1.Hence, the general solution is x(t) = c111e3t+ c21−1e−t. Michael Brannan Armin [email protected] [email protected]/4Problem 3. (10 points) The position x(t) of a certain mass on a spring is described by x′′+ cx′+ 5x = F sin (ωt).(a) Assume first that there is no extern al force , i.e. F = 0. For which values of c is the s ystem overdamped?(b) Now, F0 and the system is undamped, i.e. c = 0. For which values of ω, if any, does resonance occur?Solution.(a) The discriminant of the characteristic equation is c2−20. Hence the system is overdamped if c2−20 > 0, thatis c > 20√= 2 5√.(b) The natural freq ue nc y is 5√. Resonance therefore occurs if ω = 5√. Problem 4. (20 points) Find the general solution of the differential equation y(3)− y = ex+ 7.Solution. Let us first solve the homogeneous equation y′′′− y = 0. Its characteristic polynomial r3− 1 = (r −1)(r2+ r + 1) has roots r = 1 and r = −12±i3√2.There is a particular solution of the form yp= Axex+ B.yp′= A(x + 1)ex, yp′′= A(x + 2)ex, yp′′′= A(x + 3)exPlugging into the DE, we get yp′′′− yp= 3Aex−B7!ex+ 7. Consequently, A =13, B = −7.Hence, the general solution is −7 + (c1+13x)ex+ c2e−x/2cos3√2x+ c3e−x/2sin3√2x. Michael Brannan Armin [email protected] [email protected]/4Problem 5. (20 points) Consider, for x > 0, the second-order differential equationy′′−1 +2xy′+1x+2x2y = 0.(a) Show that the the functions y1(x) = x and y2(x) = x exare solutions to this differential equation.(b) Using the Wronskian, show that y1and y2are linearly independent solutions to the above differential equation.(c) Find, for x > 0, the general solution to the second-order differential equationy′′−1 +2xy′+1x+2x2y = 2x.Solution.(a) We havey1′′−1 +2xy1′+1x+2x2y1= 0 −1 +2x+1x+2x2x = 0,andy2′′−1 +2xy2′+1x+2x2y2= x ex+ 2ex−1 +2x(x ex+ ex) +1x+2x2(x ex) = 0.(b) The Wronskian of y1and y2is given byW (x) = dety1(x) y2(x)y1′(x) y2′(x)= detx x ex1 x ex+ ex= x2ex.Since W (x)0 on the domain of definition for the differential equation, the Wronskian theorem implies thaty1and y2are linearly independent.(c) The gene ral solution is given byy(x) = c1y1(x) + c2y2(x) + yp(x) = c1x + c2x ex+ yp(x),where ypis any pa rticular solution to the above non-homogeneous equatio n. To find such a yp, we w ill use thevariation of parameters formula:yp(x) = u1(x) y1(x) + u2(x) y2(x) = u1(x) x + u2(x) x ex,whereu1(x) = −Z2x y2(x)W (x)dx = −Z2dx = −2x,u2(x) =Z2x y1(x)W (x)dx =Z2 e−xdx = −2 e−x.So,y(x) = c1x + c2x ex−2x2−2x = d1x + d2xex−2x2. Michael Brannan Armin [email protected] [email protected]/4Problem 6. (20 points) The mo tion of a certain mass on a spring is described by x′′+ 2x′+ 2x = 5 sin (t).(a) What is the amplitude of the resulting steady periodic oscillations?(b) Assume that the mass is initially at rest (i.e. x(0) = 0, x′(0) = 0) and find the position function x(t).Solution.(a) The characteristic polynomial of the associate d ho mogeneous DE has roots−2 ± 4 − 8√2= −1 ±i.Hence, xspis the form xsp= A1cos(t) + A2sin(t).We compute xsp′= −A1sin(t) + A2cos(t) and xsp′′= −A1cos(t) −A2sin(t).Plugging into the DE gives xsp′′+ 2xsp′+ 2xsp= (A1+ 2A2)cos(t) + (A2− 2A1)sin(t)7!5sin(t). Consequ e ntly,A1+ 2A2= 0 and A2−2A1= 5, resulting in A1= −2, A2= 1.Thus, xsp= −2cos(t) + sin (t). The amplitude is (−2)2+ 12p= 5√.(b) From first part, we know that x(t) = −2cos(t) + sin (t) + e−t(c1cos (t) + c2sin (t)).Using x(0) = −2 + c1= 0 we find c1= 2.x′(t) = 2sin(t) + cos (t) −e−t(2cos(t) + c2sin (t)) + e−t(−2sin(t) + c2cos(t)). Hence, x′(0) = 1 −2 + c2= 0 resultsin c2= 1.In conclusion, x(t) = −2cos(t) + sin (t) + e−t(2 c os (t) + sin (t)). Problem 7. (5 points) Let ypbe any solution to the inhomogene ous linear differential equation y′′+ xy = ex. Finda homogeneous linear d iffer ential equation which ypsolves. Hint: Do not attempt to solve the DE.Solution. Applyddxto both sides of the differential equation to get y′′′+ xy′+ y = ex. Subtracting the two differentialequations, we get the homogeneous linear DE y′′′− y′′+ xy′+ (1 −x)y = 0. Michael Brannan Armin [email protected]