Solutions to Sample Questions (Chapter 3, Math 244)

Solutions to Sample Questions (Chapter 3, Math 244) 1. State the Existence and Uniqueness theorem for linear, second order differential equations (non-homogeneous is the most general form): SOLUTION: Let y 00 + p(t)y 0 + q(t)y = g(t), with y(t0 ) = y0 and y 0 (t0 ) = v0 . Then if p, q and g are all continuous on an open interval I containing t0 , a unique solution exists to the IVP, valid for all t in I. 2. If the W (y1 , y2 ) = t2 , can y1 , y2 be two solutions to y 00 + p(t)y 0 + q(t)y = 0? Explain. It is possible. Abel’s Theorem tells us that the Wronskian will either always be zero or never zero on the interval for which the solutions are defined- Therefore, as long as the interval does not contain the origin, they can be two solutions. 3. Construct the operator associated with the differential equation: y 0 = y 2 − 4. Is the operator linear? Show that your answer is true by using the definition of a linear operator. The operator is found by getting all terms in y to one side of the equation, everything else on the other. In this case, we have: L(y) = y 0 − y 2 This is not a linear operator. We can check using the definition: L(cy) = cy 0 − c2 y 2 6= cL(y) Furthermore, L(y1 + y2 ) = (y10 + y20 ) − (y1 + y2 )2 6= L(y1 ) + L(y2 ) 4. Find the solution to the initial value problem:    3t u + u = 3(2π − t)   0 00 if 0 ≤ t ≤ π if π < t < 2π if t ≥ 2π u(0) = 0 u0 (0) = 0 Without regards to the initial conditions, we can solve the three nonhomogeneous equations. In each case, the homogeneous part of the solution is c1 cos(t) + c2 sin(t). • u00 + u = 3t. We would start with yp = At + B. Substituting, we get: At + B = 3, so yp = 3t. The general solution in this case is: u(t) = c1 cos(t) + c2 sin(t) + 3t • u00 + u = 6π − 3t. From our previous analysis, the solution is: u(t) = c1 cos(t) + c2 sin(t) + 6π − 3t • The last part is just the homogeneous equation. The only thing left is to find c1 , c2 in each of the three cases so that the overall function u is continuous: • u(0) = 0, u0 (0) = 0 ⇒ u(t) = −3 sin(t) + 3t 0≤t≤π • u(π) = 3 and u0 (π) = 6, so: u(t) = 9 sin(t) + (3 − 6π) cos(t) + 6π − 3t π < t < 2π • u(2π) = 3 − 6π, u0 (2π) = 6: u(t) = 6 sin(t) + (3 − 6π) cos(t) 5. Solve: u00 + ω02 u = F0 cos(ωt), Undetermined Coefficients. t ≥ 2π u(0) = 0 u0 (0) = 0 if ω 6= ω0 using the Method of 6. Compute the solution to: u00 + ω02 u = F0 cos(ω0 t) u(0) = 0 u0 (0) = 0 two ways: SOLUTION: See the homework for 3.7/3.8 7. For the following question, recall that the acceleration due to gravity is 32 ft/sec2 . An 8 pound weight is attached to a spring from the ceiling. When the weight comes to rest at equilibrium, the spring has been stretched 2 feet. The damping constant for the system is 1−lb-sec/ft. If the weight is raised 6 inches above equilibrium and given an upward velocity of 1 ft/sec, find the equation of motion for the weight. Write the solution as R cos(ωt − δ), if possible. SOLUTION: First the constants. Since mg − kL = 0, we find that mg = 8, so 2k = 8, or k = 4. We are given that γ = 1, and since 8 = mg, then m = 8/32 = 1/4. 1 00 y + y 0 + 4y = 0 4 y(0) = − 1 2 y 0 (0) = −1 Or we could write: y 00 + 4y 0 + 16y = 0. Solving the characteristic equation, we get √ √ √ −4 ± 42 − 4 · 16 −4 ± 4 3 i 2 = = −2 ± 2 3 i r + 4r + 16 = 0 ⇒ r = 2 2 Therefore, the general solution is:  √  √ y(t) = e−2t C1 cos(2 3t) + C2 sin(2 3 t Solving the IVP, differentiate to get the equations for C1 , C2 :   √ √  √ √ √  y 0 = −2e−2t C1 cos(2 3t) + C2 sin(2 3 t + 2 3e−2t C2 cos(2 3t) − C1 sin(2 3 t Therefore, − 12 = C1 √ −1 = −2C1 + 2 3C2 ⇒ C1 = − 1 2 1 C2 = − √ 3 Rewrite the function using the formulas given: s y(t) = √ 7 −2t e cos(2 3t − δ) 12 √ where (−1/2, −1/ 3) is in quadrant 3, so that ! −1 δ = tan 2 √ + π ≈ 3.9987 rad 3 Maple code: b:=2*sqrt(3); G:=exp(-2*t)*(C_1*cos(b*t)+C_2*sin(b*t)); diff(G,t); Y1:=subs(C_1=-1/2, C_2=-1/sqrt(3),G); Y2:=exp(-2*t)*sqrt(7/12)*cos(2*sqrt(3)*t-d); d:=evalf(arctan(2/sqrt(3))+Pi); plot({Y1,Y2},t=0..5); 8. Given that y1 = 1 t solves the differential equation: t2 y 00 − 2y = 0 Find the fundamental set of solutions. First, rewrite the differential equation in standard form: y 00 − 2 y=0 t2 Then p(t) = 0 and W (y1 , y2 ) = Ce0 = C. On the other hand, the Wronskian is: 1 1 W (y1 , y2 ) = y20 + 2 y2 t t Put these together: 1 0 1 y2 + 2 y2 = C t t 1 y20 + y2 = Ct t The integrating factor is t, (ty2 )0 = Ct2 ty2 = C1 t3 + C2 ⇒ ⇒ C1 t2 + Notice that we have both parts of the homogeneous solution, y1 = C2 t 1 t and y2 = t2 . Alternative Solution: Use Reduction of Order, where we assume that y2 = vy1 = v t ⇒ y20 = v0t − v v0 v = − 2 2 t t t ⇒ y200 = v 00 v0 v −2 2 +2 3 t t t ⇒ y2 = C3 t2 + Substituting this back into the DE, we get tv 00 − 2v 0 = 0 ⇒ v0 = C1 t2 ⇒ v= C1 3 t + C2 3 C2 t and again we see that we can take y2 = t2 . 9. Suppose a mass of 0.01 kg is suspended from a spring, and the damping factor is γ = 0.05. If there is no external forcing, then what would the spring constant have to be in order for the system to critically damped? underdamped? The model equation can be written as: 0.01u00 + 0.05u0 + ku = 0 ⇒ u00 + 5u0 + αu = 0 where 100k = α. The solutions depend on the discriminant, 25 − 4α If this is zero, we have a system that is critically damped. In this case, k = 4/2500 If the discriminant is negative, the system is underdamped. Solving for k, we get that k > 4/2500. 10. Give the full solution, using any method(s). If there is an initial condition, solve the initial value problem. (a) y 00 + 4y 0 + 4y = t−2 e−2t Using the Variation of Parameters, yp = u1 y1 + u2 y2 , we have: y1 = e−2t y2 = te−2t g(t) = e−2t t2 with a Wronskian of e−4t . You should find that: 1 1 u01 = − u02 = 2 t t 1 t −2t −2t so yp = − ln(t)e − e . This last term is part of the homogeneous solution, so this simplifies to − ln(t)e−2t . Now that we have all the parts, u1 = − ln(t) u2 = − y(t) = e−2t (C1 + C2 t) − ln(t)e−2t (b) y 00 − 2y 0 + y = tet + 4, y(0) = 1, y 0 (0) = 1. With the Method of Undetermined Coefficients, we first get the homogeneous part of the solution, yh (t) = et (C1 + C2 t) Now we construct our ansatz (Multiplied by t after comparing to yh ): g1 = tet ⇒ yp1 = (At + B)et · t2 Substitute this into the differential equation to solve for A, B: yp1 = (At3 + Bt2 )et yp0 1 = (At3 + (3A + B)t2 + 2Bt)et yp001 = (At3 + (6A + B)t2 + (6A + 4B)t + 2B)et Forming yp001 − 2yp0 1 + yp1 = tet , we should see that A = yp1 = 61 t3 et . The next one is a lot easier! yp2 = A, so A = 4, and: 1 6 and B = 0, so that 1 y(t) = et (C1 + C2 t) + t3 et + 4 6 with y(0) = 1, C1 = −3. Solving for C2 by differentiating should give C2 = 4. The full solution:   1 3 t + 4t − 3 + 4 y(t) = et 6 (c) y 00 + 4y = 3 sin(2t), y(0) = 2, y 0 (0) = −1. The homogeneous solution is C1 cos(2t) + C2 sin(2t). Just for fun, you could try Variation of Parameters. We’ll outline the Method of Undetermined Coefficients: yp = (A sin(2t) + B cos(2t))t = At sin(2t) + Bt cos(2t) yp00 = (−4At − 4B) sin(2t) + (4A − 4Bt) cos(2t) taking yp00 + 4yp = 3 sin(2t), we see that A = 0, B = − 34 , so the solution is: 3 y = c1 cos(2t) + c2 sin(2t) − t cos(2t) 4 With y(0) = 2, c1 = 2. Differentiating to solve for c2 , we find that c2 = −1/8. 00 (d) y + 9y = N X bm cos(mπt) m=1 The homogeneous part of the solution is C1 cos(3t) + C2 sin(3t). We see that 3 6= mπ for m = 1, 2, 3, . . .. The forcing function is a sum of N functions, the mth function is: gm (t) = bm cos(mπt) ⇒ ypm = A cos(mπt) + B sin(mπt) Differentiating, yp00m = −m2 π 2 A cos(mπt) − m2 π 2 B sin(mπt) so that yp00m + 9ypm = (9 − m2 π 2 )A cos(mπt) + (9 − m2 π 2 )B sin(mπt). Solving for the coefficients, we see that A = bm /(9 − m2 π 2 ) and B = 0. Therefore, the full solution is: y(t) = C1 cos(3t) + C2 sin(3t) + N X bm cos(mπt) 2 2 m=1 9 − m π 11. Rewrite the expression in the form a + ib: (i) 2i−1 (ii) e(3−2i)t (iii) eiπ • 2i−1 = eln(2 i−1 ) = e(i−1) ln(2) = e− ln(2) ei ln(2) = 12 (cos(ln(2)) + i sin(ln(2))) • e(3−2i)t = e3t e−2ti = e3t (cos(−2t) + i sin(−2t)) = e3t (cos(2t) − i sin(2t)) (Recall that cosine is an even function, sine is an odd function). • eiπ = cos(π) + i sin(π) = −1 12. Find a linear second order differential equation with constant coefficients if y1 = 1 y2 = e−t form a fundamental set, and yp (t) = 12 t2 − t is the particular solution. The roots to the characteristic equation are r = 0 and r = −1. The characteristic equation must be r(r + 1) = 0 (or a constant multiple of that). Therefore, the differential equation is: y 00 + y 0 = 0 For yp = 21 t2 − t to be the particular solution, yp00 + yp0 = (1) + (t − 1) = t so the full differential equation must be: y 00 + y 0 = t 13. Determine the longest interval for which the IVP is certain to have a unique solution (Do not solve the IVP): t(t − 4)y 00 + 3ty 0 + 4y = 2 y(3) = 0 y 0 (3) = −1 Write in standard form: y 00 + 3 0 4 2 y + y= t−4 t(t − 4) t(t − 4) The coefficient functions are all continuous on each of three intervals: (−∞, 0), (0, 4) and (4, ∞) Since the initial time is 3, we choose the middle interval, (0, 4). 14. Let L(y) = ay 00 + by 0 + cy for some value(s) of a, b, c. If L(3e2t ) = −9e2t and L(t2 + 3t) = 5t2 + 3t − 16, what is the particular solution to: L(y) = −10t2 − 6t + 32 + e2t We see that: L(3e2t ) = −9e2t . By linearity, cL(3e2t ) = L(3ce2t ) = −9ce2t = e2t so c must be −1/9, and 1 1 L(− 3e2t ) = L(− e2t ) = e2t 9 3 Similarly, L(t2 + 3t) = 5t2 + 3t − 16 We need to multiply the right-side of the equation by −2 to get the desired part of our solution, so multiply both sides by −2: −2L(t2 + 3t) = −10t2 − 6t + 32 By linearity, −2L(t2 + 3t) = L(−2t2 − 6t). The particular solution is therefore, 1 yp (t) = −2t2 − 6t − e2t 3 15. Show that, using the substitution x = ln(t), then the differential equation: 4t2 y 00 + y = 0 becomes a differential equation with constant coefficients. Solve it. The way the ODE is written now, the derivative is with respect to t. We need to convert it to a derivative in x: dy dy dx dy 1 = · = · dt dx dt dx t And the second derivative: d2 y d = 2 dt dt Now defining y 0 = dy , dx ! dy 1 dy 1 · + − 2 dx t dx t   d2 y dx 1 dy 1 · − · = 2· dx dt t dx t2 the differential equation becomes: 2 4t  1 1 y · 2 − y 0 2 + y = 4y 00 − 4y 0 + y = 0 t t  00 The characteristic equation has solns: r = 12 , 12 y(x) = e(1/2)x (C1 + C2 x) Back substituting x = ln(t), we get: y(t) = √ t (C1 + C2 ln(t)) 16. Compute the Wronskian of two solutions of the given DE without solving it: x2 y 00 + xy 0 + (x2 − α2 )y = 0 Idea: Use Abel’s Theorem, where W (y1 , y2 ) = Ce− In this case, p(x) = x x2 R p(x) dx = x1 , so the Wronskian is W (y1 , y2 ) = C/x 17. Given t2 y 00 − 2y = 0 and y1 = 1/t, find y2 by reduction of order. Oops! This was Exercise 8. 18. If y 00 − y 0 − 6y = 0, with y(0) = 1 and y 0 (0) = α, determine the value(s) of α so that the solution tends to zero as t → ∞. The solution is: 2 + α 3t 3 − α −2t e + e 5 5 For the solution to tend to zero, the first constant must be zero, so α = −2.  y=    19. Give the general solution to y 00 + y = 1 sin(t) +t Big Idea: We can use the Method of Undetermined Coefficients on one part, and Variation of Parameters for 1/ sin(t): • For g(t) = t, we guess yp1 (t) = At + B. Substituting it back in, we get A = 1, B = 0. • For g(t) = 1 , sin(t) use Variation of Parameters, where y1 (t) = cos(t) y2 (t) = sin(t) W = cos2 t + sin2 t = 1 − sin(t) · 1/ sin(t) = −1 ⇒ u1 (t) = −t 1 cos(t) · 1/ sin(t) ⇒ u2 (t) = ln | sin(t)| u02 = 1 u01 = so yp2 = −t cos(t) + sin(t) ln | sin(t)| The general solution is y(t) = A cos(t) + B sin(t) + t − t cos(t) + sin(t) ln | sin(t)| 20. Given each of the DEs below, mark if it corresponds to beating, resonance, periodic. (a) y 00 + 36y = 5 cos(6t) RESONANCE (b) y 00 + 36y = 5 cos(5t) BEATING (c) y 00 + 36y = 0 PERIODIC 21. Describe the graph of a typical solution to: (a) y 00 + y 0 + 36y = 5 cos(5t) In this graph, we expect there to be a transient solution and a steady state solution, where the steady state solution is periodic with period 2π/5. (b) y 00 + y 0 + 36y = 0 In this graph (especially compared to the other), we have an underdamped situation, so there will be some oscillations that get smaller and smaller. (c) y 00 + 12y 0 + 36y = 0 In this graph, we have the critically damped case (no oscillations).