Short Solutions to Review Material for Test #2 MATH 3200

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1 Short Solutions to Review Material for Test # MATH 300 Kawai # Newtonian mechanics. Air resistance. a A projectile is launched vertically. Its height is y t, and y 0 = 0 and v 0 = v 0 > 0. The acceleration due to gravity is g m/s. If > 0, and the resistive force is F = v, then decide which ODE represents this case: mv = mg + v OR mv = mg v? [Recall that v is velocity.] The answer depends on the coordinate system. If y increases when you move up vertically, then acceleration should e in the NEGATIVE direction recall handout?. Since our initial velocity is upward and gravity is pointed downward, we must start with: mv = mg Now since air resistance always goes in the opposite direction of velocity, we can say: mv = mg v WHY do we need TWO initial conditions for this? If you want an IVP solution, you must provide y 0 and y 0. What if I only want to solve for v t? What type of solution previously seen is this? v + v = g m The integrating factor is µ = e /mt. [ ] e /mt v dt = g e /mt dt e /mt v = mg e/mt + C v t = mg + Ce /mt We only need the initial condition v 0 = v 0 > 0. The IVP solution is: v 0 = mg v t = mg + + C C = v 0 + mg v 0 + mg e /mt Thus, with air resistance, the velocity goes from +v 0 to mg/. So this tells us that this rilliant formula works from t = 0, still works when v changes direction, and eventually makes it to terminal velocity as long as we have availale height to go downward

2 Generally, we solve part a up until v = 0. What happens after that when v reverses direction and ecomes negative? I answered this aove. Is the solution symmetric with respect to part a? Let s look at the actual solution for y t. Integrate. y t = = mg + mg t m v 0 + mg v 0 + mg e /mt dt e /mt + C Let s assume that y 0 = 0. This gives us: 0 = m v 0 + mg + C C = m v 0 + mg The full IVP solution is: y t = mg t + m v 0 + mg e /mt Let s put in some numers: m =, g = 9.8, = /, v 0 = 0. y t = 9.6 t e t/ y x If the initial velocity is+0 m/s, then it only takes aout 3. seconds for the projectile to return to the ground. The maximum altitude is aout m. It should e clear that the trajectory is NOT symmetric with respect to time. When our velocity is positive, the air resistance is in the same direction as the downward acceleration due to gravity. When our velocity is negative, the air resistance is in the opposite direction as the downward acceleration due to gravity, and thus, the two cases are not symmetric. # nd Order Linear Constant Coeffi cient Homogeneous Spring-mass: my + y + ky = 0 Suppose m =. Give the general solutions for these: a = 0, k = 4. λ + 4 = 0 λ = ±i. Undamped. y t = c cos t + c sin t

3 = 4, k = 5. λ + 4λ + 5 = 0 λ = ± i. Underdamped. y t = e t c cos t + c sin t c = 3, k =. λ + 3λ + = 0 λ =,. Overdamped. y t = c e t + c e t #3 Solve the IVP s. a y + 9y = 0, y 0 =, y 0 = 3. y t = c cos 3t + c sin 3t y 0 = c = y t = cos 3t + c sin 3t y t = 3 sin 3t + 3c cos 3t y 0 = 3c = 3 c =. The IVP solution is y t = cos 3t + sin 3t. [As in the HW, I could ask: Find the first positive value of t where y = 0.] This occurs when: cos 3t sin 3t cos 3t = sin 3t = tan 3t = cos 3t cos 3t The first opportunity occurs in Quadrant II: 3t = 3π 4 t = π 4. y + 5y + 6y = 0, y 0 =, y 0 = 5. y t = c e 3t + c e t y 0 = c + c = Oh, great. eqns. in unknowns. y t = 3c e 3t c e t y 0 = 3c c = 5 Multiply the first one y 3. Eliminate. 3c + 3c = 6 3c c = 5 If we add them, we otain c = and then we mus thave c = 9. y t = 9e 3t + e t 3

4 #4 Find the particular solution, y p t. a y + y + y = 9e t. Left side: λ + λ + = 0 λ =,. Right side: λ =. No conflicts! The particular solution must e y p = Ae t. Sustitute and alance. The particular solution is y p t = e t. The general solution has the c s in it: y p = Ae t, y p = 4Ae t 4Ae t + Ae t + Ae t = 9e t 9Ae t = 9e t A =. y t = e t c + c t + e t. y + y = 3 cos t. Yuck. unknown coeffi cients. Left side: λ + = 0 λ = ±i. Right side: λ ± i. No conflicts! Sustitute and alance. y p = A cos t + B sin t y p = 4A cos t 4B sin t 4A cos t 4B sin t + A cos t + B sin t = 3 cos t 3A cos t 3B sin t = 3 cos t We must have A = and B = 0. The particular solutions is: y p t = y p = cos t #5 Give the form of the particular solution do not solve ecause these would take nearly forever. a y + 4y = t sin t + t Left side: λ + 4 = 0 λ = ±i. Right side: λ = ±i, ±i, ±i, 0, 0. One conflict. y p = At + Bt + Ct 3 cos t + Dt + Et + F t 3 sin t + G + Ht We must promote each cosine & sine solution y step. y + y + y = e t cos t Left side: λ + λ + = 0 λ = ± i. Right side: λ = ± i. Same. Promote. y p = te t A cos t + B sin t 4

5 #6 Find the Wronskian for the independent solutions for: a y + y = 0 λ 3 + λ = 0 λ = 0, ±i. {cos t, sin t, } cos t sin t 0 sin t cos t 0 cos t sin t = sin t cos t cos t sin t = sin t cos t =. If you haven t expanded minors, then don t worry. I just wrote this one for the Calc. 3 folks... y + 9y = 0 λ = ±3i. {cos 3t, sin 3t} cos 3t sin 3t 3 sin 3t 3 cos 3t = 3 cos 3t 3 sin 3t = 3. c y + 4y + 4y = 0 λ =,. e t e t te t t + e t #7 Perform the Variation of Parameters on: { e t, te t} a We have p t = 0, so we can just let 0dt = C. = t + e 4t te 4t = e 4t y + y = sec t y = cos t, y = sin t The Wronskian is equal to from aove. sec t sin t v = dt = tan t dt = ln cos t. Assume that cos t > 0. The particular solution is: sec t cos t v = dt = t. y p = ln cos t cos t + t sin t y y + y = et t, y = e t, y = te t The Wronskian is e t te t e t t + e t = et. This gives us: e t /t te t v = e t dt = dt = t. e t /t e t dt v = e t dt = = ln t. t This second one only works ecause the exponentials cancel! 5

6 The particuular solution is: y p = t e t + ln t te t??? The first term is part of the homogeneous solution, so it does not count. The est answer is: y p = te t ln t. I sustituted this in, and it worked! #8 Evaluate the operators: a D + td [ e 3t] = [ e 3t] + t [ e 3t ] = 9e 3t + 3te 3t. What is the annihilator of e t? D c What is the annihilator of t 3 e t? D 4. Four times! d What is the annihilator of sin t? D + 4 e What is the annihilator of [cos t + sin t]? Same. cosine & sine. #9 Sect. 4.7, Exercises #4 & #44. Reduction of Order. D + 4 will annihilate oth #4 If you already read aout Cauchy-Euler equations, then you proaly already know the other solution. 4 y y t t y = 0, y = t We have: The numerator integral is: e y = t ptdt t dt p t = t e lnt = t Thus, we have: #44 We have: Same: Thus, we have: t t dt = t t t y + y + t t t 4 dt = t 5 = t4 5 5 = y t t y = 0, y = e t t p t = = t t et lnt = et t e t e t /t e t dt = et t dt = et ln t = y This only works ecause the e t in the numerator cancels with the y in the denominator. 6

7 { #0 Using the definition of LT, explain why we do not have L e t}. Laplace Transform. The definition is: L {f t} = 0 e st f t dt The function e t is super-exponential, so: lim t e t, for any positive value of s. est Thus, the Laplace Transform integral will not converge. # Evaluate L {f t}. a f t = t + = t + 4t + 4 f t = sin t/3 L {sin t/3} = c f t = e t cos 5t a =, = 5. L { t + 4t + 4 } = s s + 4 s /3 s + /3 = 9 /3 9 s + /9 = 3 9s + L { e t cos 5t } = s + s d f t = t 5 u t 5 The unshifted function is t and a = 5. { } L t 5 u t 5 = e 5s s 3 e f t = t 3 u t 7 I proaly should NOT have chosen t 3, ecause now I have to do this: g t = t 3 g t + 7 = t = t 3 + t + 47t Thus, the correct answer is: e 7s L { t 3 + t + 47t } 6 = e 7s s s s s f f t = sin t u t g t = sin t g t + = sin t + = cos sin t + sin cos t The correct answer is: e s cos s s + sin + s + 7

8 # Evaluate L {y } if y 0 =, y 0 =, and y 0 = 3. #3 Evaluate L {F s}. a F s = /s 5. F s = / s 4 s 3 Y s s y 0 sy 0 y 0 = s 3 Y s s s 3 { } { } 4 s L s 5 = t 4 L 5 = 4 t4 { } { } L 6 s 4 = t 3 e t L s 4 = 6 t3 e t c F s = 5s + 6 s + 9 { } s L s + 9 { } 3 = cos 3t and L s = sin 3t + 9 Thus, we must have: { 5s L s } s = 5 cos 3t + sin 3t. + 9 d F s = s 4 s. Complete the square. 8s + 7 { } L s 4 s 4 = e 4t cos t + e F s = e s s 5. Right shift is. Original function was f t = /4 t 4. f t u t = 4 t 4 u t f F s = e s s. Right shift is. Original function was f t = / sin t. + 4 f t u t = sin t u t #4 Evaluate: { 5s L 3 + 5s } + s + s s =??? + Partial fraction: Multiply through y LCD. 5s 3 + 5s + s + s s + = A s + B s + Cs + D s + 5s 3 + 5s + s + = As s + + B s + + Cs + D s 8

9 s = 0: We have: Balance s : We have: Balance s: We have: Balance s 3 : We have: = B 5s 3 + 5s + s + = As s + + s + + Cs + D s 5s = s + Ds D = 3 5s 3 + 5s + s + = As s + + s + + Cs + 3 s s = As A = 5s 3 + 5s + s + = s s + + s + + Cs + 3 s 5s 3 = s 3 + Cs 3 C = 4 { L s + s + 4s + 3 } s = + t + 4 cos t + 3 sin t + Too long for an exam, ut it is good practice. #5 Suppose we are solving something simple with Laplace Transform: y = g t We know that Laplace will always give us a continuous solution, as long as we provide appropriate initial conditions at t = 0. a If g t looks like this: 4 y Write g t in terms of the unit step u t. t + t u t. 3 Assume that y 0 = 0. Solve this with Laplace t Is your solution differentiale at t =? YES. Fundamental Thm. of Calculus says: if g t is continuous, then: G t = t 0 g s ds is continuous AND G t = g t. Antiderivative property. Solving: t = t Thus, the unit step term is: t t u t and this is properly shifted one to the right. 9

10 The original function is f t = t, and then it was shifted to the right. sy s y 0 = s e s s Y s = s 3 e s s 3 The inverse LT is: y t = { }! L s + { }! L e s s + = t t u t What does this look like? On [0,, we have y t = / t, the paraola. On [,, the unit step part gets turned on, and we have: t t = t. The solution is continuous, and it is a SPLINE ecause the first derivative is continuous and curve is smooth no sharp turns!. y t Suppose g t looks like this instead: Write g t in terms of the unit step u t. 4 y t + t u t = t t u t + u t There is an extra shifted term t Solve this with Laplace. Is your solution differentiale at t =? NO. What is different at t =? We have a sharp turn. The original function is f t = t, and then it was shifted to the right. sy s y 0 = s e s s + e s sy s = s s s e s + e s s Y s = s 3 e s s 3 + e s s 0

11 The inverse LT is: { }! y t = L s + { } { }!! L e s s + + L e s s + = t t u t + t u t What does this look like? On [0,, we have y t = / t, the paraola. On [,, the Heaviside part gets turned on, and we have: t t + t = t 3. The solution is continuous, ut we have a sharp turn at t =. y t #6 Unit step: #7 Solve: a Find L { e s/ /s }. This matches: e as s, a = { } L e s/ /s Find L {sin t u t }. This matches: f t u t e s F s = u t. The original function is f t = sin t, so we have: L {sin t u t } = e s s. + y 3y + y = t u t, y 0 = 0, y 0 =. This is proaly a it too long for an exam, ut it is almost reasonale. Left side: L { y } = s Y s sy 0 y 0 = s Y s 3L { y } = 3 sy s y 0 = 3sY s L {y} = Y s

12 This gives us: Right side: L { y 3y + y } = s 3s + Y s L {t u t } = e s L {t} = e s s. Both sides, together: s 3s + Y s = e s s Y s = We consult our note sheet for the partial fractions. a =, = : s s = s s + e s s s s = + s s s s The partial fractions for the last one require 4 unknown coeffi cients. s s s = A s + B s + C s + D s s = 0 B = /. s = C =. s = D = /4. = As s s + B s s + Cs s + Ds s If we look at the coeffi ent of s 3 on oth sides, we see that the left side has coeffi ent 0 and the right side has coeffi cient: 0 = A + C + D = A + + A = Thus, we have Y s: Y s = s + 3 s + e s 4 + s s + s 4 s 3 y t = e t + e t t et + 4 et h t Rememer that the Inverse LT of the stuff after the e s term is: t et + 4 et, ut we must SHIFT one unit to the right. Constant functions stay constant after the shift.

13 #8 Suppose: g t = h t h t 3 + h t 5 h t What might this look like? Use geometric series and find F s. L {g t} = e s e 3s + e 5s e 7s +... s The common ratio is r = e s. S = a r = s e s e s = e s + e s #9 Suppose on [0, 4, g t = e t. Now turn that function off at t = 4, and turn on g t = the shifted paraola y = t 4 units to the right. Find L {g t}. We need on off. g t = e t + t 4 e t h t 4 = e t + t 4 h t 4 e t h t 4 The last term is tricky. This is NOT in the correct form: f t 4 h t 4,. so we must rewrite it. e t = e t 4 e 8 Thus, we have: g t = e t + t 4 h t 4 e 8 e t 4 h t 4 L {g t} = s + e 4s! s 3 e 8 e 4s s = s + e 4s s 3 e 4s. s 3

14 More Practice! #0 Solve: y y = t t u t + t u t y y = t + t u t y y = t t u t Note my transformations. We now have something that looks like f t u t. Let s say that we only care aout the solution on [0, ]. We will not other with turning off f t at t =. Here is a graph of f t. y x In this case we have f t = t and we have: f t = t. Good. L { t} = /s and, we have: L { t u t } = e s s We need to make this into an IVP: a LT of Left Side & Right Side Initial condition: y 0 = sy s Y S = s + e s s Move.& factor: s Y s = s + e s s c Divide: Y s = s s + s e s s s d Partial fractions. Fortunately for us, we only need to solve: I did the wolframalpha.com thing... e Inverse LT: For the first part, we have: s s = A s + B s + C s = s s + s L { s s + s 4 } = t + e t

15 The second part is a shift: { L e s s s + } = t + e t s = t + e t u t We take the same functions of t from the previous part, ut sustitute t for t. f Check for continuity? When t =, you should e adding zero. + e = + = 0. It might still e wrong, ut at least you know you created a continuous function. g Final answer: # Find L {F s}. a Partial fractions: Multiply through y the LCD. y t = t + e t t + e t u t 4s + 3s + s s + s + 3 = A s + B s + + C s + 3 4s + 3s + = A s + s Bs s Cs s + Fortunately, there are 3 critical numers, so we will get A, B, and C quickly. s = 0: = A 3 A =. s = : s = 3: = B = B B = = C = 3C C = 3. Thus, we have partial fractions and the Inverse LT. } L { + 3 = e t + 3e 3t. s s + s + 3 The other one: s 6 s = A s + B s If we do this y partial fractions, it is still easy. s 6 = A s + B There is only critical numer. s = : 6 = B = 4. 5

16 So we have: s 4 = A s 4 Either sustitute another value in for s [s = 0 is a good choice], or alance the coeffi cients with the polynomial thm. The left side coeffi cient of s is. The right side coeffi cient of s is A. Thus, we must have A =. L { 4 s The first one is easy. The second one looks like: } s F s = s with F s = L { e t t } Thus, the final answer is: e t + te t. 6