Calculus IV - HW 3 Due 7/13 Section 3.1 1. Give the general solution to the following differential equations: a y 25y = 0 Solution: The characteristic equation is r 2 25 = r 5r + 5. It follows that the general solution is y = c 1 e 5t + c 2 e 5t. b y + 6y + 8y = 0 Solution: The characteristic equation is r 2 +6r+8 = r+2r+4. It follows that the general solution is y = c 1 e 2t + c 2 e 4t. c y + 12y = 0. Solution: The characteristic equation is r 2 +12r = rr 2 +12. It follows that the general solution is y = c 1 e 0t + c 2 e 12t = c 1 + c 2 e 12t. 2. For each equation from problem 1; assume the constants in your general solution, φt, are both positive, and find lim φt and lim φt. t t 1
Solution: a lim φt = lim c 1 e 5t + c 2 e 5t = t t lim φt = lim c 1e 5t + c 2 e 5t = t t b lim φt = lim y = c 1 e 2t + c 2 e 4t = 0 t t lim φt = lim y = c 1e 2t + c 2 e 4t = t t c lim φt = lim y = c 1 + c 2 e 12t = c 1 t t lim φt = lim y = c 1 + c 2 e 12t = t t Section 3.2 3. 7,11,12 from B&D For each of the following differential equations with initial conditions, determine the longest interval in which a solution is guaranteed to exist. a ty + 3y = t; y1 = 1, y 1 = 2 Solution: In order to apply the existence and uniqueness theorem for second order linear differential equations we need to put the equation in the form y + pty + qty = gt. Some manipulation of our equation yields y + 3 t y = 1. Our task now is to find the largest open interval containing the t value at which our initial condition is specified on which both functions qt = 3 t and gt = 1 are continuous. g is continuous everywhere and q is continuous for t 0. This gives us the option of two intervals:, 0 and 0,. Choosing the one containing 1 where the initial condition is specified leaves us with a final answer of 0,. b x 3y + xy + ln x y = 0; y1 = 0, y 1 = 1 Page 2
Solution: Some manipulation of our equation yields y + x ln x x 3 y + x 3 y = 0. Now we need to find the largest open interval containing the x value at which our initial condition is specified on which both functions pt = and ln x x 3 gt = are continuous. p is continuous for x 3 and q is continuous for x 0, 3. This leaves us with three possible intervals:, 0, 0, 3, and 3,. Choosing the one containing 1 where the initial condition is specified leaves us with a final answer of x x 3 0, 3. c x 2y + y + x 2 tanxy = 0; y3 = 1, y 3 = 2 Solution: Some manipulation of our equation yields y + 1 x 2 y + tanxy = 0. Now we need to find the largest open interval containing the x value at which our initial condition is specified on which both functions pt = 1 and x 2 gt = tanx are continuous. p is continuous for x 2 and q is continuous for x π + kπ with k an integer. This leaves us with infinitely many possible 2 intervals. We only need to find the one containing 3. The final answer is 2, 3π 2. 4. Optional 41 from B&D The equation P xy +Qxy +Rxy = 0 is said to be exact if it can be written in the form [pxy ] +[fxy] = 0, where fx is to be determined in terms of P x, Qx, and Rx. The latter equation can be integrated once immediately, resulting in a first order linear equation. By equating the coefficients of the preceding equations and then eliminating fx, show that a necessary condition for exactness is P x Q x + Rx = 0 meaning exactness implies P x Q x + Rx = 0. It can be shown that this is also a sufficient condition. Solution: Page 3
Proof. To prove this we first assume that P xy + Qxy + Rxy = 0 is exact. In particular, this means that P xy + Qxy + Rxy = [pxy ] + [fxy]. Expanding the right hand side of the above equation we have Rearranging we have It follows that [pxy ] + [fxy] = p xy + pxy + f xy + fxy. pxy + [p x + fx]y + f xy. P xy + Qxy + Rxy = pxy + [p x + fx]y + f xy. Then by equating coefficients we have P x = px, and In particular, and Qx = p x + fx, Rx = f x. Q x = p x + f x = P x + Rx P x + Rx Q x = P x Q x + Rx = 0. 5. Problems 4,6 from B&D Find the Wronskian of each of the following pairs of functions. a x, xe x b cos 2 θ, 1 + cos2θ Solution: a W = x xex 1 1 + xe x = x1 + xex xe x = x 2 e x Page 4
b W = cos 2 θ 1 + cos2θ 2 cosθ sinθ 2 sin2θ = cos 2 θ 1 + cos2θ 2 cosθ sinθ 4 cosθ sinθ = 4 cos 3 θ sinθ 1 + cos2θ 2 cosθ sinθ = 4 cos 3 θ sinθ 2 cos 2 θ 2 cosθ sinθ = 4 cos 3 θ sinθ + 4 cos 3 θ sinθ = 0 Note that we used the identities sin2θ = 2 sinθ cosθ and cos 2 θ = 1 2 + cos2θ 2. 6. Problems 8,9,10 from B&D For each of the following differential equations with initial conditions, determine the longest interval on which the given initial value problem is certain to have a unique twice differentiable solutions. a t 1y 3ty + 4y = sint y 2 = 2, y 2 = 1 b y + costy + 3 ln t y = 0 y2 = 3 y 2 = 1 Solution: a We need to divide through by t 1 to put the equation into the standard form. Then we are left with y 3t t 1 y + 4 t 1 y = sint t 1. Then thinking of pt = 3t, qt = 4 sint, and gt =, we see that t 1 t 1 t 1 pt, qt, and gt are all continuous everywhere except at t = 1. Thus our two intervals to choose between are I 1 : < t < 1 and I 2 : 1 < t <. However we need this interval to contain which in this case is = 2. Therefore we want to pick I 1. b Here pt = cost, qt = 3 ln t, and gt = 0. We see that pt and gt are continuous everywhere. However qt has a discontinuity at t = 0. Thus we must choose between the intervals I 1 = < t < 0 and I 2 = 0 < t <. Now our = 2 so we choose I 2. Page 5
7. Problems a and b are 24 and 27 from B&D For each of the following, verify that the functions y 1 and y 2 are solutions of the given differential equation and state whether or not they constitute a fundamental set of solutions. a y + 4y = 0; y 1 t = cos2t y 2 t = sin2t b 1 x cotx y xy + y = 0 y 1 x = x, y 2 x = sinx c t 2 y 2ty + 2y = 0 y 1 t = t 2 y 2 t = t 2 Solution: a We note that y 1 = 4 cos2t and y 2 = 4 sin2t so so y 1 and y 2 are both solutions. y 1 + 4y 1 = 4 cos2t + 4 cos2t = 0 y 2 + 4y 2 = 4 sin2t + 4 sin2t = 0 We have shown that two solutions constitute a fundamental set of solutions if their Wronskian is not identically zero, meaning it is not zero everywhere. Thus we compute W = cos2t sin2t 2 sin2t 2 cos2t = 2 cos2 2t + 2 sin 2 2t = 2 Since W 0, we have that y 1 and y 2 constitute a fundamental set. b Note that y 1 = 1 and y 1 = 0 while y 2 = cosx and y 2 = sinx. Plugging these in we get 1 x cotx y 1 xy 1 + y 1 = 1 x cotx0 x1 + x = 0 x + x = 0 and 1 x cotx y 2 xy 2 + y 2 = 1 x cotx sinx x cosx + sinx = sinx + x cosx x cosx + sinx = 0 All that remains is to check their Wronkskian. W = x sinx 1 cosx = x cosx sinx which in particular is nonzero at x = 2π. Thus y 1 and y 2 form a fundamental set of solutions. Page 6
c We will demonstrate that y 1 is a solution then note that since y 2 is a constant multiple of y 1, it must also be a solution. Note y 1 = t 2, y 1 = 2t, and y 1 = 2. Plugging this in we have t 2 y 1 2ty 1 + 2y 1 = t 2 2 2t2t + 2t 2 = 0 We then compute the Wronskian. W = t2 t 2 2t 2t = 2t3 + 2t 3 = 0 so y 1 and y 2 do not form a fundamental set. Section 3.3 8. Use Euler s Formula to write the following in the form a + ib. a e 2 3i b e iπ Solution: a We use Euler s Formula to calculate e 2 3i = e 2 e 3i = e 2 cos 3 + i sin 3 = e 2 cos3 i sin3 = e 2 cos3 ie 2 sin3. b Similarly e iπ = cosπ + i sinπ = 1 + i0 = 1. 9. Problems 7,9,12 from B&D Find the general solution of the following differential equations. a y 2y + 2y = 0 b y + 2y 8y = 0 c 4y + 9y = 0 Page 7
Solution: a We get the characteristic equation r 2 2r + 2 = 0. Using the quadratic formula, we find that the roots are 2± 4 42 2 = 1 ± i. However, we have shown that if our roots are complex, namely r = λ ± iµ, then our general equation is just y = c 1 e λt sin µt + c 2 e λt cosµt. Here λ = 1 and µ = 1 so we have the general solution y = c 1 e t sint + c 2 e t cost. b Our characteristic equation is r 2 + 2r 8 = 0 which can be factored to r 2r + 4 so it has two real roots, r = 2 and r = 4. Thus its general solution is just y = c 1 e 2t + c 2 e 4t. c The characteristic equation is 4r 2 +9 = 0 which implies that the solutions are r = ± 3 i. Following the example of parta, we get that the general solution 2 is 3 3 y = c 1 cos 2 t + c 2 sin 2 t. Note that λ = 0 in this example. 10. Problem 17 and 22 from B&D Find the solutions of the following initial value problems. Sketch the graph of each solution and describe its behavior as t. a y + 4y = 0, y0 = 0, y 0 = 1 b y + 2y + 2y = 0 y π 4 = 2, y π 4 = 2 Solution: a Note that we showed in Problem 3a that y = c 1 sin2t + c 2 cos2t is a general solution. Thus it only remains to determine the appropriate c 1 and c 2. We get y0 = c 1 sin0 + c 2 cos0 = c 2 = 0 y 0 = 2c 1 cos0 2c 2 sin0 = 2c 1 = 1 so c 1 = 1 and the solution of the initial value problem is y = 1 sin2t. Thus 2 2 yt oscillates with fixed amplitude and frequency as t. Page 8
b We find the characteristic equation r 2 + 2r + 2 = 0. We use the quadratic formula to get r = 2± 4 4 2 2 = 1 ± i. Thus our general solution is We calculate y = c 1 e t sint + c 2 e t cost. y t = c 1 e t sint c 2 e t cost + c 1 e t cost c 2 e t sint = c 1 c 2 e t sint + c 1 c 2 e t cost. Then π y 4 = c 1 e π π 4 sin + c 2 e π π 4 cos 4 4 = 2 2 c 1 + c 2 e π 4 = 2 = c 1 + c 2 = 2 2e π 4. Furthermore π y = c 1 c 2 e π π 4 sin + c 1 c 2 e π π 4 cos 4 4 4 2 π 2 = 4 c 1 c 2 + c 1 c 2 = 4 2c2 2 e = c 2 2e π 4 = 2 = c 2 = 2e π 4. π 2 e However we showed earlier that c 1 + c 2 = 2 2e π 4 = c 1 = 2 2e π 4 c 2 so c 1 = 2 2e π 4 2e π 4 = 2e π 4. Therefore the solution to the initial value problem is y = 2e π 4 e t sint + 2e π 4 t cost = 2e π 4 t sint + 2e π 4 e t cost. Thus yt has a decaying oscillation as t. Page 9
This graph has been drastically vertically stretched in a non-uniform manner to better illustrate the general trend of the graph. 11. Optional Problem 28 of Section 3.3 from B&D In this problem we outline a different derivation of Euler s formula. a Show that y 1 t = cost and y 2 t = sint are a fundamental set of solutions of y + y = 0; that is, show they are solutions and that their Wronskian is not zero. b Show formally that y = e it is also a solution. Why does this imply that for some constants c 1 and c 2? c Set t = 0 in equation 1 to show that c 1 = 1. e it = c 1 cost + c 2 sint 1 d Differentiate equation 1 and set t = 0 to show that c 2 = i. It may be helpful to remember that d dt ert = re rt for any r C. Use the values of c 1 and c 2 to arrive at Euler s Formula. Solution: a Since y 1t = cost and y 2t = sint, it follows immediately that these are solutions. We check and see W = cost sint sint cost = cos2 t + sin 2 t = 1 so since W 0, it follows that y 1 and y 2 are a fundamental set of solutions. Page 10
b If y = e it, then y = ie it and y = i 2 e it = e it. Thus y + y = e it + e it = 0 meaning y = e it is a solution. However, by the definition of y 1 and y 2 being a fundamental set, it is possible to write any other solution as c 1 y 1 + c 2 y 2 for some c 1, c 2. Thus there exist c 1 and c 2 such that e it = c 1 cost + c 2 sint. c Setting t = 0 we get. e i0 = c 1 cos0 + c 2 sint = 1 = c 1 d Taking the derivative of equation 1, we get ie it = c 1 sint + c 2 cost which if we plug in t = 0 gives i = c 2. Thus we find e it = cost + i sint which is Euler s Formula. Section 3.4 12. Find the general solution of the following differential equation. 2.25y + 3y + y = 0 Solution: The characteristic equation is 2.25r 2 + 3r + 1 = 2 3 2 r + 1 = 0 which has the repeated root r = 2. It follows that the general solution is 3 y = c 1 e 2t/3 + c 2 te 2t/3. 13. Find the solution of the following differential equation with initial conditions. 4y 20y + 25y = 0; y1 = e 5/2, y 1 = 7e5/2 2 Page 11
Solution: The characteristic equation is 4r 2 20r + 25 = 2r 5 2 = 0 which has the repeated root r = 5. Hence the general solution is 2 y = c 1 e 5t/2 + c 2 te 5t/2. Plugging in the initial conditions we have y1 = c 1 + c 2 e 5/2 = e 5/2 implying that and implying that y 1 = c 1 + c 2 = 1, 2 5 2 c 1 + 7 2 c 2 e 5/2 = 7 2 e5/2 5c 1 + 7c 2 = 7. 3 Solving the system of two equations comprised of 1 and 2 we have that c 1 = 0 and c 2 = 1. This leaves us with at final solution of y = te 5t/2. 14. Suppose that we have the differential equation ay + by + cy = 0. a Write the characteristic equation and use the quadratic formula to find its roots in terms of a, b, and c. b Recall that the discriminant is b 2 4ac. Note that there are three possibilities: i. b 2 4ac > 0 implying there are two distinct real roots ii. b 2 4ac = 0 implying there is one repeated real root. iii. b 2 4ac < 0 implying there are two distinct complex roots. For each of these cases, give a formula for the general solution that only involves real-valued functions meaning they don t include imaginary numbers. Page 12
Solution: a The characteristic equation is ar 2 + br + c = 0 and it has roots r 1 = b + b 2 4ac 2a ; r 2 = b b 2 4ac. 2a b i. When b 2 4ac > 0, we have two distinct real roots and we learned in Section 3.1 that y = c 1 e b+ b 2 4ac 2a t + c 2 e b b 2 4ac t 2a is a general solution. ii. When b 2 4ac = 0, we have one repeated root and we saw in Section 3.4 that a general solution is given by y = c 1 e b 2a t + c 2 te b 2a t. iii. When b 2 4ac < 0, we have two distinct complex roots. We showed in Section 3.3 that a general solution is given by y = c 1 e b 4ac b 2a t 2 sin t + c 2 e b 4ac b 2a t 2 cos t 2a 2a 15. Use your work in Problem 14 to write a general solution for each of the following homogeneous differential equations. a 9y + 9y 4y = 0 b y 5y + 6y = 0 c y + 8y + 16y = 0 Solution: a Here a = 4, b = 0, and c = 9. Thus b 2 4ac = 449 = 144 < 0. In particular b = 0 and 4ac b 2 = 144 = 12 = 3. Thus our general solution is 2a 2a 2 4 8 2 3 3 y = c 1 sin 2 t + c 2 cos 2 t. Page 13
b Here a = 1, b = 5, and c = 6. Thus b 2 4ac = 25 416 = 1 > 0. Then have we have the general solution y = c 1 e b+ b 2 4ac 2a t + c 2 e b b 2 4ac t 2a = c 1 e 5+ 25 416 2 t + c 2 e 5 25 416 2 t = c 1 e 3t + c 2 e 2t c Here a = 1, b = 4, and 4. Thus b 2 4ac = 4 2 414 = 0. Then have we have the general solution Section 3.5 y = c 1 e b 2a t + c 2 te b 2a t = c 1 e 2t + c 2 te 2t 16. Find the general solution of the following differential equation. y + 8y + 25y = 40 cos5t Solution: We begin by finding the solution to the associated homogeneous equation, y + 8y + 25y = 0. This has a characteristic equation of with roots r 2 + 8r + 25 = 0 r = 4 ± 3i. It follows that the general solution to the associated homogeneous equation is e 4t c 1 cos3t + c 2 sin3t. Now we just need to find a particular solution to the nonhomogeneous case. We do this with the method of undetermined coefficients. Let y = A cos5t + B sin5t and notice that y = 5A sin5t + 5B cos5t, Page 14
y = 25A cos5t 25B sin5t. Plugging into the equation we get 25A cos5t 25B sin5t+8 5A sin5t+5b cos5t+25a cos5t+b sin5t = 40A sin5t + 40B cos5t = 40 cos5t. This implies that A = 0 and B = 1. Hence we have a particular solution of y = sin5t and a general solution of y = e 4t c 1 cos3t + c 2 sin3t + sin5t. 17. Find the general solution of the following differential equations. a 9y + 12y + 4y = 18e t/3 Solution: We begin by finding the general solution to the associated homogeneous case. The characteristic equation is 9r 2 + 12r + 4 = 3r + 2 2 = 0 which has the repeated root r = 2. It follows that the general solution to 3 the associated homogeneous case is y = c 1 e 2t/3 + c 2 te 2t/3. We now find a general solution using the method of undetermined coefficients. Let y = Ae t/3 and notice then that Plugging into the equation we have y = A 3 et/3, y = A 9 et/3. Ae t/3 + 4Ae t/3 + 4Ae t/3 = 9Ae t/3 = 18e t/3. It follows that A = 2 and we have a general solution of y = c 1 e 2t/3 + c 2 te 2t/3 + 2e t/3. Page 15
b 9y + 12y + 4y = 25 cost/3 Solution: Since we already have the general solution to the associated homogeneous case, we only need to find a particular solution to the nonhomogeneous case. Letting y = A cost/3 + B sint/3 we have Plugging into the equation we have y = A 3 sint/3 + B 3 cost/3, y = A 9 cost/3 B 9 sint/3. A cost/3 B sint/3 4A sint/3+4b cost/3+4a cost/3+4b sint/3 = 4B + 3A cost/3 + 3B 4A sint/3 = 25 cost/3 Equating coefficients we have the system of equations 4B + 3A = 25 3B 4A = 0. Solving this we have B = 4 and A = 3. It follows that we have a particular solution of y = 3 cost/3 + 4 sint/3, and a general solution of y = c 1 e 2t/3 + c 2 te 2t/3 + 3 cost/3 + 4 sint/3. c 9y + 12y + 4y = 18e t/3 + 25 cost/3 Hint Consider the principle of superposition. Solution: Treating the left hand side of the equation as a differential operator, it follows that the general solution in this case is y = c 1 e 2t/3 + c 2 te 2t/3 + 2e t/3 + 3 cost/3 + 4 sint/3. 18. Solve the following differential equation using the two methods described, and then solve for the values of the constants. y 2y = 6t 2 6t 12; y1 = 1 2, y 1 = 1 a i. Let v = y and solve the resulting first order linear equation. Page 16
Solution: Making the substitution we have the equation v 2v = 6t 2 6t 12 which we can solve as a first order linear equation. Using the integrating factor µ = e 2t we have µv = e 2t 6t 2 6t 12 dt = 1 2 e 2t 6t 2 6t 12 + 1 2 e 2t 12t 6 dt [ 1 2 e 2t 12t 6 + 1 2 ] 12e 2t dt = 1 2 e 2t 6t 2 6t 12 + 1 2 = e 2t 3t 2 + 3t + 6 e 2t 3t + 3 3 2 2 e 2t + c 1 = e 2t 3t 2 + 6 + c 1. Dividing through by µ = e 2t we have v = 3t 2 + 6 + c 1 e 2t. ii. Notice that y = v dt and use this to find the general solution. Solution: We have that y = v dt = 3t 2 + 6 + c 1 e 2t dt = t 3 + 6t + c 1 e 2t + c 2. b i. Solve the homogeneous case using the roots of the characteristic equation. Solution: The characteristic equation of the associated homogeneous equation is r 2 2r = rr 2 = 0. It follows that the general solution to the homogeneous equation is y = c 1 e 2t + c 2. Page 17
ii. Referencing your solution from part a, make a guess for a general form of a particular solution and use the method of undetermined coefficients to find the general solution. Solution: From the above it is clear that the a particular solution to this differential equation is y = t 3 + 6t. This is a bit odd because, based on the methods we have developed, we would expect the particular solution to be a quadratic. The reason that we need a cubic is that there is no y term on the left hand side of the equation hence if we plugged in a general quadratic polynomial we the resulting polynomial would be linear. So, playing the game that the question is asking us to play, let and notice then that Plugging into the equation we have y = At 3 + Bt 2 + Ct y = 3At 2 + 2Bt + C, y = 6At + 2B. 6At + 2B 23At 2 + 2Bt + C = 6At 2 + 6A 4Bt 2C = 6t 2 6t 12 It follows that A = 1, B = 0, and C = 6. Hence we have the particular solution y = t 3 + 6t as expected. Again we find the general solution of y = t 3 + 6t + c 1 e 2t + c 2. c Solve for the constants in your general solution using the given initial values. Solution: Plugging in the initial values we have and y1 = 5 + c 1 e 2 + c 2 = 1 2 y 1 = 3 + 2c 1 e 2 = 1. It follows that c 1 = e 2t and c 2 = 7. The solution is then 2 y = t 3 + 6t e 2t 2 7 2. Page 18
19. 28 from B&D Optional Determine the general solution of N y + λ 2 y = a m sinmπt, m=1 where λ > 0 and λ mπ for m = 1,..., N. Solution: We begin by solving for the general solution of the associated homogeneous equation y + λ 2 y = 0. The characteristic equation is r 2 + λ 2 = 0 which has roots r = ±iλ. It follows that the general solution to the homogeneous equation is y = c 1 sinλt + c 2 cosλt. Treating the left hand side of the differential equation as a linear operator, we will find a particular solution to y + λ 2 y = a m sinmπt and then sum over m. It turns out, in this case, that we only need to plug A m sinmπt in to solve for a particular solution. We then have Plugging into our equation we have It follows that y = A m sinmπt, y = m 2 π 2 A m sinmπt. m 2 π 2 A m sinmπt + λ 2 A m sinmπt = a m sinmπt A m = a m [λ 2 m 2 π 2 ] 1. Summing over m and adding in the general solution to the associated homogeneous equation we have the general solution y = c 1 sinλt + c 2 cosλt + N a m [λ 2 m 2 π 2 ] 1 sinmπt. m=1 20. Use the method of undetermined coefficients to find a particular solution to each of the following nonhomogeneous differential equations. Page 19
a 9y + 9y 4y = 3e t + sin2t b y 5y + 6y = t 2 4t + 2 c y + 4y + 4y = 4e t cost Solution: a We will find a particular solution Y t to 4y + 9y = 3e t + sin2t by finding particular solutions Y 1 t and Y 2 t to 4y + 9y = 3e t and 4y + 9y = sin2t respectively. For Y 1 t, our guess is Ae t. Then Y 1 t = Ae t Y 1t = Ae t Y 1 t = Ae t Thus plugging these into the differential equation gives 4Ae t + 9Ae t = 3e t = 13A = 3 = A = 3 13 so we have Y 1 = 3 13 e t. For Y 2 our guess is A sin2t + B cos2t. Thus Plugging these in we have Y 1 t = A sin2t + B cos2t Y 1t = 2A cost 2B sint Y 1 t = 4A sint + 4B cost 4 4A sint + 4B cost + 9 A sint + B cost = sin2t = 7A sint = sin2t and 7B cos2t = 0 cos2t = 7A = 1 and 7B = 0 = A = 1 7 and B = 0. Therefore we have Y 2 = 1 7 sin2t. Thus Y t = Y 1 t + Y 2 t = 3 13 e t 1 7 sin2t. Page 20
b Our guess is Y t = At 2 + Bt + C so we have Y t = At 2 + Bt + C Y t = 2At + B Y t = 2A Plugging this in we get 2A 5 2At + B + 6 At 2 + Bt + C = t 2 4t + 2 = 6At 2 + 6B 10At + 2A 5B + 6C = t 2 4t + 2 = 6A = 1, 6B 10A = 4, 2A 5B + 6C = 2 = A = 1 6 Plugging this in we get 6B 10A = 4 = 6B 10 1 6 = 4 Finally we get = B = 14 36 2A 5B + 6C = 2 1 6 514 36 + 6C = 2 = C = 5 108 Thus the solution is Y t = 1 6 t2 14t 5. 36 108 c Here our guess is Y t = Ae t cost + Be t sint so we have Y t = Y t = Ae t cost + Be t sint Y t = A + Be t cost + B Ae t sint e t cost + A + B + B A = 2Be t cost 2Ae t sint B A A + B e t sint Thus plugging in we get 2Be t cost 2Ae t sint + 4 A + Be t cost + B Ae t sint Page 21
+4 Ae t cost + Be t sint = 4e t cost Collecting terms, we get 2B + 4A + 4B + 4A e t cost = 8A + 6B e t cost = 4e t cost 2A + 4B 4A + 4B e t sint = 6A + 8B e t sint = 0e t sint Thus 6A + 8B = 0 which implies B = 3 A. Substituting this in we get 4 4 = 8A + 18A which implies 50 16 A = 4 so A = = 8 and it follows that 4 4 50 25 B = 3A = 24 = 6. 4 100 25 Therefore the solution is Y t = 8 25 et cost + 6 25 et sint. 21. Use your work in Problem 14 and Problem 20 to find a general solution to the following nonhomogeneous differential equations. a 9y + 9y 4y = 3e t + sin2t b y 5y + 6y = t 2 4t + 2 c y + 4y + 4y = 4e t cost Solution: Recall that we get the general solution to a nonhomogeneous equation by adding together the general solution to the homogeneous solution and a particular solution to the nonhomogenenous equation. However, for each of these, we found the general solution to the homogeneous equation in Problem 2 and a particular solution the nonhomogeneous equation in Problem 3. Thus all that remains is to add them together. a yt = c 1 sin 3 2 t + c 2 cos 3 2 t + 3 13 e t 1 7 sin2t b yt = c 1 e 3t + c 2 e 2t + 1 6 t2 14 36 t 5 108 c yt = c 1 e 2t + c 2 te 2t + 8 25 et cost + 6 25 et sint Section 3.6 22. Parts i-iv are optional. Parts v-vii are mandatory. This question explores the general method of variation of parameters. Suppose we have a second order linear differential equation in standard form, y + pty + qty = gt, Page 22
and the general solution to the homogeneous case, y = c 1 y 1 + c 2 y 2. We attempt to find the general solution of the non-homogeneous case by the following means: a replacing the constants, c 1 and c 2, with functions of t, u 1 t and u 2 t b stipulating that u 1y 1 + u 2y 2 = 0 c plugging the resulting expression into the non-homogeneous differential equation d and solving for u 1 and u 2. Preform the steps below to derive a general solution method. i. Let y = u 1 y 1 + u 2 y 2 and compute expressions for y and y bearing in mind the condition given in b above. Hint u 1 and u 2 should not appear in any of these expressions. ii. Using your results from i, plug y into the non-homogeneous differential equation given at the onset of this problem. iii. Rearrange the left hand side of the equation you found in part ii so that the equation is in the form u 1 [ ] + u 2 [ ] + u 1[ ] + u 2[ ] = gt where each instance of [ ] represents an expression in y, y, y, p and q; all four of these will be different expressions. iv. Explain why the expressions from iii that are multiplied by u 1 and u 2 are zero and notice that this leaves us with the equation u 1y 1 + u 2y 2 = gt. v. Using the equations from iv and b we have the following system of linear not differential equations for u 1 and u 2. u 1y 1 + u 2y 2 = 0 1 u 1y 1 + u 2y 2 = g 2 Solve the system using the following steps: A. Multiply equation 1 by y 2 and equation 2 by y 2 B. Add the resulting equations together and solve for u 1 C. Multiply equation 1 by y 1 and equation 2 by y 1 D. Add the resulting equations together and solve for u 2. Page 23
vi. Recall that that the Wronskian of y 1 and y 2 is W y 1, y 2 = y 1 y 2 y 1 y 2 = y 1y 2 y 1y 2 and show that your result from v. can be rewritten as u 1 = y 2g W y 1, y 2, u 2 = y 1 g W y 1, y 2. vii. Find an expression for the general solution of the non-homogeneous equation the one we ve been trying to solve this whole time in terms of the indefinite integrals of the expressions for u 1 and u 2 from part vi. Argue that this is the general solution. Hint remember the constants of integration! Solution: i. If y = u 1 y 1 + u 2 y 2, the product rules gives us that y = u 1y 1 + u 1 y 1 + u 2y 2 + u 2 y 2 = u 1 y 1 + u 2 y 2 + u 1y 1 + u 2y 2 = y = u 1 y 1 + u 2 y 2 with the last equivalence coming from the stipulation that u 1y 1 + u 2y 2 = 0. Then ii. We then get y = u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 u 1 y 1 + u 1y 1 + u 2 y 2 + u 2y 2 + ptu 1 y 1 + u 2 y 2 + qtu 1 y 1 + u 2 y 2 = gt iii. Separating out terms, we get u 1 y 1 + pty 1 + qty 1 + u 2 y 2 + pty 2 + qty 2 + u 1y 1 + u 2 y 2 = gt. iv. Both y 1 and y 2 are solutions of y + pty + qty = gt so and the equation becomes y 1 + pty 1 + qty 1 = 0 = y 2 + pty 2 + qty 2 u 1y 1 + u 2y 2 = gt. Page 24
v. A. After multiplying, we get the system of equations u 1y 1 y 2 u 2y 2 y 2 = 0 u 1y 1y 2 + u 2y 2 y 2 = y 2 g B. Adding these together, we get u 1 y 1y 2 y 1 y 2 = y 2 g Thus u 1 = y 2 g y 1y 2 y 1 y 2 C. After multiplying, we get the system of equations u 1y 1 y 1 u 2y 2 y 1 = 0 u 1y 1y 1 + u 2y 2y 1 = y 1 g D. Adding these together, we get u 2 y 1 y 2 y 1y 2 = y 1 g Solving for u 2 we get u 2 = y 1 g y 1 y 2 y 1y 2. vi. This is immediate from the definition of the Wronskian. vii. Recall that we set out to find y = u 1 y 1 + u 2 y 2. To get u 1 and u 2, we simply integrate u 1 and u 2. Thus we have y 2 sgs y = y 1 W y 1, y 2 s ds + y y 1 sgs 2 W y 1, y 2 s ds. Remember that each of the indefinite integrals will have constant of integration, c 1 and c 2 respectively so the equation could equally well be written y 2 sgs y = y 1 W y 1, y 2 s ds + y y 1 sgs 2 W y 1, y 2 s ds + c 1y 1 + c 2 y 2 y which is the general solution since Y 1 t = y 2 sgs 1 ds+y W y 1,y 2 s 2 is a particular solution to the nonhomogeneous equation and Y 2 t = c 1 y 1 + c 2 y 2 is a general solution to the homogeneous equation. y 1 sgs W y 1,y 2 s ds Page 25
23. 5 from B&D Find the general solution of the following differential equation. y + y = tant, t 0, π/2 Solution: First we must find a fundamental set of solutions to the homogeneous equation y + y = 0. We see the characteristic equation is r 2 + 1 = 0 so r = ±i implying the general solution is y = c 1 sint + c 2 cost. Since this is a general solution, any choice of c 1 and c 2 gives a solution. Choosing c 1 = 0 and c 2 = 1 gives a solution y 1 = cost. Choosing c 1 = 1 and c 2 = 0 gives a solution y 2 = sint. Since we have two solutions, it only remains to see if they form a fundamental set. Taking the Wronskian, we get W = cost sint sint cost = cos2 t + sin 2 t = 1 0 so y 1 and y 2 do form a fundamental set. Note that since tant is not continuous everywhere, we must use the definite integral form of the theorem see Theorem 3.6.1 on p. 188 of B&D. We already found that W y 1, y 2 = 1. Note that is strictly inside the interval 0, π/2. We get y = cost sins tansds + sint + We now need to find these integrals. coss tansds + c 1 cost + c 2 sint. sin 2 s t coss ds = 1 cos 2 s t ds = secs cossds coss sins tansds = t = ln sec s+tan s sins = ln sec t+tan t sint ln sec +tan +sin. Thus noting that ln sec + tan + sin is just some constant C 1, we have that Next, we calculate sins tansds = ln sec t + tan t sint + C 1. coss tansds = t sinsds = coss = cost cos Page 26
= cost + C 2 Putting this all back together, we get y = cost ln sec t+tan t sint+c 1 +sint cost+c 2 +c 1 cost+c 2 sint = cost ln sec t + tan t + sint cost C 1 cost sint cost + C 2 sint +c 1 cost + c 2 sint = cost ln sec t + tan t + C 2 + c 2 sint + c 1 C 1 cost. = cost ln sec t + tan t + D 1 cost + D 2 sint. 24. 10 from B&D Find the general solution of the following differential equation. y 2y + y = et 1 + t 2 Solution: As before, we begin by finding the general solution to the homogeneous equation y 2y + y = 0 which has characteristic equation r 2 2r+1 = 0 which has the repeated root r = 1. Thus the general solution is y = c 1 e t + c 2 te t. Thus we could take our two solutions to be y 1 = e t and y 2 = te t. We check the Wronskian and see W = et te t e t t + 1e t = t + 1e2t te 2t = e 2t 0 so y 1 and y 2 do in fact form a fundamental set. Since each part is continuous on the whole real line, we may simply use the indefinite integral form to get et et e y = e t 1+t 2 et t dt + te t 1+t 2 ds + c 1 e t + c 2 te t = e t e 2t t dt + tet 1 + t2 = e t 1 2 ln 1 + t2 + C 1 e 2t 1 1 + t 2 dt + c 1e t + c 2 te t + te t arctan t + C 2 + c 1 e t + c 2 te t = 1 2 et ln 1 + t 2 + te t arctant + c 1 C 1e t + c 2 + C 2 te t = 1 2 et ln 1 + t 2 + te t arctant + D 1 e t + D 2 te t. Page 27
25. Find the general solution to the following differential equation. y + y = sec 4 t, t 0, π/2 Hint sec 3 t dt = 1 sect tant + 1 ln sect + tant + c. 2 2 Solution: Looking at Problem 2, we get that y 1 = cost and y 2 = sint form a fundamental set of solutions with W y 1, y 2 = 1. Since sect has discontinuities, we use the definite integral form of variation of parameters to get that y = cost sins sec 4 sds + sint coss sec 4 sds + c 1 sint + c 2 cost We calculate the integrals below: sins sec 4 sds = tans sec 3 sds. We now use u-substitution with u = secs and du = secs tansds to write the integral as u 2 du = 1 3 u3 = 1 3 sec3 s Thus we have [ 1 cost sins sec 4 sds = cost 3 sec3 s t ] = 1 3 cost sec3 t + 1 3 cost sec3 = 1 3 sec2 t + C 1 cost For the other integral we have coss sec 4 sds = sec 3 sds = 1 2 secs tans + 1 2 ln secs + tans t Page 28
= 1 2 sect tant + 1 ln sect + tant 2 1 2 sec tan + 1 2 ln sec + tan = 1 2 sect tant + 1 2 ln sect + tant + C 2 Thus sint coss sec 4 sds = 1 2 sint sect tant + 1 2 sint ln sect + tant + C 2 sint = 1 2 tan2 t + 1 2 sint ln sect + tant + C 2 sint Putting this all together we get y = 1 3 sec2 t + 1 2 tan2 t + 1 2 sint ln sect + tant + C 1 cost + C 2 sint Page 29