Applied Differential Equations 2280Sample Final ExamWednesday, 6 May 2009, 7:30-10:15amInstructions: This in-class exam is 50 minutes. No calculators, notes, tables or books.No answer check is expected. Details count 75%. The answer counts 25%.1. (Quadrature Equation)Solve for the general solution y(x) in the equation y0= 2 cot x+1250x31 + 25x2+x ln(1+x2).[The required integration talent includes basic formulae, integration by parts, substitutionand college algebra.]Answer:y = 2 ln (sin (x)) +492x2− ln1 + 25 x2+ 1/21 + x2ln1 + x2− 1/2 + C2. (Separable Equation Test)The problem y0= f(x, y) is said to be separable provided f(x, y) = F (x)G(y) forsome functions F and G.(a) [75%] Check ( X ) the problems that can be put into separable form, but don’tsupply any details.y0= −y(2xy + 1) + (2x + 3)y2yy0= xy2+ 5x2yy0= ex+y+ ey3y0+ 5y = 10y2(b) [25%] State a test which can verify that an equation is not separable. Use thetest to verify that y0= x +q|xy| is not separable.Answer:(a) yy0= xy2+ 5x2y is not separable, but the other three are separable.(b) Test: fy/f not independent of x implies not separable.Let f = x +q|xy| and assume x > 0. Then y > 0 and f = x +√xy. We have fy= 1/yand fy/f = y/(x +√xy) depends on x, so the DE is not separable.3. (Solve a Separable Equation)Given y2y0=2x2+ 3x1 + x212564− y3.(a) Find all equilibrium solutions.(b) Find the non-equilibrium solution in implicit form.To save time, do not solve for y explicitly.Answer:(a) y = 5/4(b)−13ln |125 − 64y3| = 2x +32ln(1 + x2) − 2 arctan(x) + c4. (Linear Equations)(a) [60%] Solve 2v0(t) = −32 +23t + 1v(t), v(0) = −8. Show all integrating factorsteps.(b) [30%] Solve 2√x + 2dydx= y. The answer contains symbol c.(c) [10%] The problem 2√x + 2 y0= y − 5 can be solved using the answer yhfrompart (b) plus superposition y = yh+yp. Find yp. Hint: If you cannot write the answerin a few seconds, then return here after finishing all problems on the exam.Answer:(a) v(t) = −24t − 8(b) y(x) = Ce√x+25. (Stability)(a) [50%] Draw a phase line diagram for the differential equationdx/dt = 10002 −5√x3(2 + 3x)(9x2− 4)8.Expected in the diagram are equilibrium points and signs of x0(or flow directionmarkers < and >).(b) [40%] Draw a phase diagram using the phase line diagram of (a). Add these labelsas appropriate: funnel, spout, node, source, sink, stable, unstable. Show at least 8threaded curves. A direction field is not expected or required.(c) [10%] Outline how to solve for non-equilibrium solutions, without doing any inte-grations or long details.Answer:(a) and (b) See a handwritten exam solution for a similar problem on midterm 1.(c) Put the DE into the form y0/G(y) = F (x) and then apply the method of quadrature.6. (ch3)(a) Solve for the general solutions:(a.1) [25%] y00+ 4y0+ 4y = 0,(a.2) [25%] yvi+ 4yiv= 0,(a.3) [25%] Char. eq. r(r − 3)(r3− 9r)2(r2+ 4)3= 0.(b) Given 6x00(t) + 7x0(t) + 2x(t) = 0, which represents a damped spring-mass systemwith m = 6, c = 7, k = 2, solve the differential equation [15%] and classify theanswer as over-damped, critically damped or under-damped [5%]. Illustrate in aphysical model drawing the meaning of constants m, c, k [5%].Answer:(a)1: r2+ 4r + 4 = 0, y = c1y1+ c2y2, y1= e−2x, y2= xe−2x.2: riv+ 4r2= 0, roots r = 0, 0, 2i, −2i. Then y = c1e0x+ c2xe0x+ c3cos 2x + c4sin 2x.3: Write as r3(r − 3)3(r + 3)2(r2+ 4)3= 0. Then y is a linear combination of theatoms 1, x, x2, e3x, xe3x, x2e3x, e−3x, xe−3x, cos 2x, x cos 2x, x2cos 2x, sin 2x, x sin 2x,x2sin 2x.Part (b)Use 6r2+ 7r + 2 = 0 and the quadratic formula to obtain roots r = −1/2, −2/3. Thenx(t) = c1e−t/2+c2e−2t/3. This is over-damped. The illustration shows a spring, dampenerand mass with labels k, c, m, x and the equilibrium position of the mass.7. (ch3)Determine for yvi+ yiv= x + 2x2+ x3+ e−x+ x sin x the shortest trial solutionfor ypaccording to the method of undetermined coefficients. Do not evaluate theundetermined coefficients!Answer:The homogeneous solution is a linear combination of the atoms 1, x, x2, x3, cos x, sin xbecause the characteristic polynomial has roots 0, 0, 0, 0, i, −i.1 An initial trial solution y is constructed for atoms 1, x, e3x, e−3x, cos x, sin x givingy = y1+ y2+ y3+ y4,y1= d1+ d2x + d3x2+ d4x3,y2= d5cos x + d6x cos x,y3= d7sin x + d8x sin x,y4= d9e−x.Linear combinations of the listed independent atoms are supposed to reproduce, by spe-cialization of constants, all derivatives of the right side of the differential equation.2 The correction rule is applied individually to each of y1, y2, y3, y4.The result is the shortest trial solutiony = y1+ y2+ y3+ y4,y1= d1x4+ d2x5+ d3x6+ d4x7,y2= d5x cos x + d6x2cos x,y3= d7x sin x + d8x2sin x,y4= d9e−x.Some facts:• The number of terms in each of y1to y4is unchanged.• If an atom of the homogeneous equation appears in a group, then it is removed.The crossed-out term is replaced by adding another term on the end of that group.• Suppose a group has base atom A. The number s of crossed-out terms for thisgroup is exactly the number of atoms of the homogeneous equation having baseatom A.This number s is the corresponding root multiplicity for base atom A in the charac-teristic equation. The value s appears in the Edwards-Penney table containing themystery factor xs.8. (ch3)(a) [50%] Find by undetermined coefficients the steady-state periodic solution for theequation x00+ 4x0+ 6x = 10 cos(2t).(b) [50%] Find by variation of parameters a particular solution ypfor the equationy00+ 3y0+ 2y = xe2x.Answer:(a) Undetermined coefficients for x00+ 4x0+ 6x = 10 cos(2t).We solve x00+ 4x0+ 6x = 0. The characteristic equation roots are −2 ±√2i and theatoms are x1= e−2tcos√2t, x2= e−2tsin√2t.The trial solution is computed from f = 10 cos 2t. We find all atoms in the derivativelist f, f0, f00. . . , then take a linear combination to form the trial solution x = d1cos 2t +d2sin 2t. No corrections are needed, because none of these terms are solutions of thehomogeneous equation x00+ 4x0+ 6x = 0.Substitute the trial solution to obtain the answers d1= 5/17, d2= 20/17. Cramer’s rulewas used to solve for the unknowns d1, d2.The unique periodic solution xssis extracted from the general solution x = xh+ xpbycrossing out all negative exponential terms (terms which limit to zero at infinity). Becausexh(t) contains atoms x1, x2(see above) containing a negative exponential factor, thenthe steady state