Task compilation - Differential Equation II
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The document contains solutions to differential equation problems. It solves 5 problems involving determining characteristic equations and solutions for differential equations of varying orders. The solutions involve finding real or complex eigenvalues and expressing the solutions as combinations of exponential and trigonometric functions with arbitrary constants.
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Task compilation - Differential Equation II
- 1. DIFFERENTIAL EQUATION II TASK COMPILATION 3/27/2014 MARIA PRISCILLYA PASARIBU IDN. 4103312018
- 2. TASK I (February, 13th 2014) Determine the solution of:: 1. 𝑑2 𝑦 𝑑𝑡2 − 2 𝑑𝑦 𝑑𝑡 − 5𝑦 = 0, y(0)=0; y’(0)=1 2. 𝑑2 𝑦 𝑑𝑡2 − 3 𝑑𝑦 𝑑𝑡 = 0, y(0)=0; y’(0)=1 3. 𝑑2 𝑦 𝑑𝑡2 − 4 𝑑𝑦 𝑑𝑡 − 4𝑦 = 0, y(0)=0; y’(0)=1 Solution: 1. 𝑑2 𝑦 𝑑𝑡2 − 2 𝑑𝑦 𝑑𝑡 − 5𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation: 𝜆2 − 2𝜆 − 5 = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 = −(−2) ± √(−2)2 − 4(1)(−5) 2(1) = 2 ± √4 + 20 2 = 2 ± √24 2 = 2 ± 2√6 2 = 1 ± √6 𝜆1and 𝜆2 are real and distinct, then 𝑦 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = −𝑐2 .................... (1) 𝑦′(0) = 1 𝑦′(𝑥) = (1 + √6)𝑐1 𝑒(1+√6)𝑥 + (1 − √6)𝑐1 𝑒(1−√6)𝑥 1 = (1 + √6)𝑐1 + (1 − √6)𝑐2 1 = (1 + √6)𝑐1 + (1 − √6)(−𝑐1) 1 = 𝑐1(1 + √6 − 1 + √6) 1 = 2√6𝑐1 𝑐1 = 1 12 √6
- 3. 𝑐1 = −𝑐2 .................... (1) 𝑐2 = − 1 12 √6 Maka, 𝑦 = 1 12 √6𝑒(1+√6)𝑥 − 1 12 √6𝑒(1−√6)𝑥 2. 𝑑2 𝑦 𝑑𝑡2 − 3 𝑑𝑦 𝑑𝑡 = 0, y(0)=0; y’(0)=1 Characteristic Equation:𝜆2 − 3𝜆 = 0 𝜆(𝜆 − 3) = 0 𝜆1 = 0 ∨ 𝜆2 = 3 𝜆1and 𝜆2 are real and distinct, then 𝑦 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 𝑦 = 𝑐1 𝑒0𝑥 + 𝑐2 𝑒3𝑥 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = −𝑐2 .................... (1) 𝑦′(0) = 1 𝑦′(𝑥) = 3𝑐2 3𝑥 1 = 3𝑐2 𝑐2 = 1 3 𝑐1 = − 1 3 Then, 𝑦 = − 1 3 𝑒 𝑚1 𝑥 + 1 3 𝑒 𝑚2 𝑥 3. 𝑑2 𝑦 𝑑𝑡2 − 4 𝑑𝑦 𝑑𝑡 − 4𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation:𝜆2 − 4𝜆 − 4 = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 = −(−4) ± √(−4)2 − 4(1)(−4) 2(1) = 4 ± √16 + 16 2 = 4 ± 4√2 2 = 2 ± 2√2
- 4. 𝜆1and 𝜆2 are real and distinct, then 𝑦 = 𝑐1 𝑒 𝑚1 𝑥 + 𝑐2 𝑒 𝑚2 𝑥 𝑦 = 𝑐1 𝑒(2+2√2)𝑥 + 𝑐2 𝑒(2−2√2)𝑥 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = −𝑐2 .................... (1) 𝑦′(0) = 1 𝑦′(𝑥) = (2 + 2√2)𝑐1 𝑒(2+2√2)𝑥 + (2 − 2√2)𝑐2 𝑒(2−2√2)𝑥 1 = (2 + 2√2)𝑐1 + (2 − 2√2)𝑐2 1 = (2 + 2√2)𝑐1 + (2 − 2√2)(−𝑐1) 1 = 𝑐1(2 + 2√2 − 2 + 2√2) 1 = 𝑐1(4√2) 𝑐1 = 1 8 √2 𝑐1 = −𝑐2 .................... (1) 𝑐2 = − 1 8 √2 Maka, 𝑦 = 1 8 √2𝑒(2+2√2)𝑥 − 1 8 √2𝑒(2−2√2)𝑥
- 5. TASK II (February, 20th 2014) Solve these equation: 1. 𝑑4 𝑦 𝑑𝑥4 + 10 𝑑2 𝑦 𝑑𝑥2 + 9𝑦 = 0 2. 𝑑4 𝑦 𝑑𝑥4 + 𝑑3 𝑦 𝑑𝑥3 + 𝑑2 𝑦 𝑑𝑥2 + 2𝑦 = 0 3. 𝑦′′′ + 4𝑦′ = 0 4. 𝑦(4) + 4𝑦′′ − 𝑦′ + 6𝑦 = 0 5. 𝑑6 𝑦 𝑑𝑥6 − 4 𝑑5 𝑦 𝑑𝑥5 + 16 𝑑4 𝑦 𝑑𝑥4 − 12 𝑑3 𝑦 𝑑𝑥3 + 41 𝑑2 𝑦 𝑑𝑥2 − 8 𝑑𝑦 𝑑𝑥 + 26𝑦 = 0 Solution: 1. 𝑑4 𝑦 𝑑𝑥4 + 10 𝑑2 𝑦 𝑑𝑥2 + 9𝑦 = 0 Characteristic equation: 𝜆4 + 10𝜆2 + 9 = 0 (𝜆2 + 9)(𝜆2 + 1) = 0 (𝜆 + 3𝑖)(𝜆 − 3𝑖)(𝜆 + 𝑖)(𝜆 − 𝑖) = 0 𝜆1 = −3𝑖 ⋁ 𝜆2 = −3𝑖 ⋁ 𝜆3 = −𝑖 ⋁ 𝜆3 = 𝑖 So, the solution is 𝑦 = 𝐶1 cos 3𝑥 + 𝐶2 sin 3𝑥 + 𝐶3 cos 𝑥 + 𝐶4 sin 𝑥 2. 𝑑4 𝑦 𝑑𝑥4 + 𝑑3 𝑦 𝑑𝑥3 + 𝑑2 𝑦 𝑑𝑥2 + 2𝑦 = 0 Characteristic equation: 𝜆4 + 𝜆3 + 𝜆2 + 2 = 0 (𝜆2 − 𝜆 + 1)(𝜆2 + 2𝜆 + 2) = 0 (𝜆2 − 𝜆 + 1) = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆1,2 = −(−1) ± √(−1)2 − 4(1)(1) 2(1) 𝜆1,2 = 1 ± √1 − 4 2 𝜆1,2 = 1 ± √3𝑖 2 𝜆1 = 1 2 + 1 2 √3𝑖 and 𝜆2 = 1 2 − 1 2 √3𝑖 (𝜆2 + 2𝜆 + 2) = 0 𝜆3,4 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎
- 6. 𝜆3,4 = −(2) ± √(2)2 − 4(1)(2) 2(1) 𝜆3,4 = −2 ± √4 − 8 2 𝜆3,4 = −2 ± 2𝑖 2 𝜆3 = −1 + 𝑖 and 𝜆4 = −1 − 𝑖 So, the solution is 𝑦 = 𝑒 1 2 𝑥 (𝐶1 cos 1 2 √3𝑥 + 𝐶2 sin 1 2 √3𝑥) + 𝑒−𝑥 (𝐶3 cos 𝑥 + 𝐶4 sin 𝑥) 3. 𝑦′′′ + 4𝑦′ = 0 Characteristic equation: 𝜆3 + 4𝜆 = 0 𝜆(𝜆2 + 4) = 0 𝜆(𝜆 + 2𝑖)(𝜆 − 2𝑖) = 0 𝜆1 = 0 ⋁ 𝜆2 = −2𝑖 ⋁ 𝜆3 = 2𝑖 So, the solution is 𝑦 = 𝐶1 + 𝐶2cos2𝑥 + 𝐶3 sin 2𝑥 4. 𝑦(4) + 4𝑦′′ − 𝑦′ + 6𝑦 = 0 Characteristics equation is 𝜆4 + 4𝜆2 − 𝜆 + 6 = 0 (𝜆2 − 𝜆 + 2)(𝜆2 + 𝜆 + 3) = 0 (𝜆2 − 𝜆 + 2) = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆1,2 = −(−1) ± √(−1)2 − 4(1)(2) 2(1) 𝜆1,2 = 1 ± √1 − 8 2 𝜆1,2 = 1 ± √7𝑖 2 𝜆1 = 1 2 + 1 2 √7𝑖 and 𝜆2 = 1 2 − 1 2 √7𝑖 (𝜆2 + 𝜆 + 3) = 0 𝜆3,4 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆3,4 = −(1) ± √(1)2 − 4(1)(3) 2(1)
- 7. 𝜆3,4 = −1 ± √1 − 12 2 𝜆3,4 = −1 ± √11𝑖 2 𝜆3 = − 1 2 + 1 2 √11𝑖 and 𝜆4 = − 1 2 − 1 2 √11𝑖 So, the equation is: 𝑦 = 𝑒 1 2 𝑥 (𝐶1 cos 1 2 √7𝑥 + 𝐶2 sin 1 2 √7𝑥) + 𝑒− 1 2 𝑥 (𝐶3 cos 1 2 √11 𝑥 + 𝐶4 sin 1 2 √11 𝑥) 5. 𝑑6 𝑦 𝑑𝑥6 − 4 𝑑5 𝑦 𝑑𝑥5 + 16 𝑑4 𝑦 𝑑𝑥4 − 12 𝑑3 𝑦 𝑑𝑥3 + 41 𝑑2 𝑦 𝑑𝑥2 − 8 𝑑𝑦 𝑑𝑥 + 26𝑦 = 0 Characteristic equation is 𝜆6 − 4𝜆5 + 16𝜆4 − 12𝜆3 + 41𝜆2 − 8𝜆 + 26 = 0 (𝜆4 + 3𝜆2 + 2)(𝜆2 − 4𝜆 + 13) = 0 (𝜆2 + 1)(𝜆2 + 2)(𝜆2 − 4𝜆 + 13) = 0 (𝜆2 + 1) = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆1,2 = 0 ± √0 − 4(1)(1) 2(1) 𝜆1,2 = ±√−4 2 𝜆1,2 = ±2𝑖 2 𝜆1 = 𝑖 and 𝜆2 = −𝑖 (𝜆2 + 2) = 0 𝜆3,4 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆3,4 = 0 ± √0 − 4(1)(2) 2(1)
- 8. 𝜆3,4 = ±√−8 2 𝜆3,4 = ±2√2𝑖 2 𝜆3 = √2𝑖 and 𝜆4 = −√2𝑖 (𝜆2 − 4𝜆 + 13) = 0 𝜆5,6 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆5,6 = −(−4) ± √(−4)2 − 4(1)(13) 2(1) 𝜆5,6 = 4 ± √16 − 52 2 𝜆5,6 = 4 ± 6𝑖 2 𝜆5 = 2 + 3𝑖 and 𝜆6 = 2 − 3𝑖 So, the solution is 𝑦 = 𝐶1 cos 𝑥 + 𝐶1 sin 𝑥 + 𝑒2𝑥(𝐶3 cos 3𝑥 + 𝐶4 sin 3𝑥) + 𝐶5 cos √2𝑥 + 𝐶5 sin √2𝑥
- 9. TASK III (March 7th, 2014) 1. 𝑑3 𝑞 𝑑𝑡3 − 5 𝑑2 𝑞 𝑑𝑡2 + 25 𝑑𝑞 𝑑𝑡 − 125𝑞 = −60𝑒7𝑡 y(0)=0, y’(0)=1, y”(0)=2 2. 𝑦(𝐼𝑉) − 6𝑦′′′ + 16𝑦′′ + 54𝑦′ − 225𝑦 = 100𝑒−2𝑥 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Solution: 1. 𝑑3 𝑞 𝑑𝑡3 − 5 𝑑2 𝑞 𝑑𝑡2 + 25 𝑑𝑞 𝑑𝑡 − 125𝑞 = −60𝑒7𝑡 y(0)=0, y’(0)=1, y”(0)=2 Quadratic equation: 𝜆3 − 5𝜆2 + 25𝜆 − 125 = −60𝑒7𝑡 Y= yl + yr 𝜆3 − 5𝜆2 + 25𝜆 − 125 = 0 (𝜆 − 5)(𝜆 + 5𝑖)(𝜆 − 5𝑖) = 0 𝑦𝑙 = 𝑐1 𝑒5𝑡 + 𝑐2 cos 5𝑡 + 𝑐3 sin 5𝑡 𝑦𝑟 = 𝐴0 𝑒7𝑡 𝑦𝑟 ′ = 7𝐴0 𝑒7𝑡 𝑦𝑟 ′′ = 49𝐴0 𝑒7𝑡 𝑦𝑟 ′′′ = 343𝐴0 𝑒7𝑡 𝑦′′′ − 5𝑦′′ + 25𝑦 − 125𝑦 = −60𝑒7𝑡 343𝐴0 𝑒7𝑡 − 245𝐴0 𝑒7𝑡 + 175𝐴0 𝑒7𝑡 − 125𝐴0 𝑒7𝑡 = −60𝑒7𝑡 148𝐴0 𝑒7𝑡 = −60𝑒7𝑡 𝐴0 = − 60 148 = − 15 37 𝑦𝑟 = − 15 37 𝑒7𝑡
- 10. Then, we got the equation is equal to Y= yl + yr 𝑦(𝑡) = 𝑐1 𝑒5𝑡 + 𝑐2 cos 5𝑡 + 𝑐3 sin 5𝑡 − 15 37 𝑒7𝑡 Now, we are going to find the value of c1, c2, and c3. y(0)=0 0 = 𝑐1 + 𝑐2 − 15 37 𝑐1 + 𝑐2 = 15 37 ........................................ (1) y’(0)=1 𝑦′ (𝑡) = 5𝑐1 𝑒5𝑡 − 5𝑐2 sin 5𝑡 + 5𝑐3 cos 5𝑡 − 105 37 𝑒7𝑡 1 = 5𝑐1 + 5𝑐3 − 105 37 𝑐1 + 𝑐3 = 142 185 .................................... (2) y”(0)=2 𝑦′′ (𝑡) = 25𝑐1 𝑒5𝑡 − 25𝑐2 cos 5𝑡 − 25𝑐3 sin 5𝑡 − 735 37 𝑒7𝑡 2 = 25𝑐1 − 25𝑐2 − 735 37 25𝑐1 − 25𝑐2 = 809 37 𝑐1 − 𝑐2 = 809 925 ....................................... (3) Elimination of 𝑐2 from (1) and (3) 𝑐1 − 𝑐2 = 809 925 ....................................... (3) 𝑐1 + 𝑐2 = 15 37 ........................................ (1) 2𝑐1 = 809 + 375 925 𝑐1 = 592 925 From (3), we can get the value of c2 by substituting the value of c1. 𝑐1 − 𝑐2 = 809 925 ....................................... (3) +
- 11. 592 925 − 𝑐2 = 809 925 𝑐2 = − 217 925 From (2), we can get the value of c3 by substituting the value of c1. 𝑐1 + 𝑐3 = 142 185 .................................... (2) 592 925 + 𝑐3 = 142 185 𝑐3 = 118 925 After getting the value of c1, c2, and c3, then the equation is: 𝑦(𝑡) = 592 925 𝑒5𝑡 − 217 925 cos 5𝑡 + 118 925 sin 5𝑡 − 15 37 𝑒7𝑡 2. 𝑦(𝐼𝑉) − 6𝑦′′′ + 16𝑦′′ + 54𝑦′ − 225𝑦 = 100𝑒−2𝑥 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Quadratic equation: 𝜆4 − 6𝜆3 + 16𝜆2 + 54𝜆 − 225 = 100𝑒−2𝑥 𝑦 = 𝑦𝑙 + 𝑦𝑟 𝜆4 − 6𝜆3 + 16𝜆2 + 54𝜆 − 225 = 0 By using Horner method and ABC formula, so that the roots are gotten: 𝜆1 = 3 𝜆3 = 3 + 4𝑖 𝜆2 = −3 𝜆4 = 3 − 4𝑖 𝑦𝑙 = 𝑐1 𝑒3𝑡 + 𝑐2 𝑒−3𝑡 + 𝑒3𝑡 (𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) 𝑦𝑟 = 𝐴0 𝑒−2𝑥 𝑦𝑟 ′ = −2𝐴0 𝑒−2𝑡
- 12. 𝑦𝑟 ′′ = 4𝐴0 𝑒−2𝑡 𝑦𝑟 ′′′ = −8𝐴0 𝑒−2𝑡 𝑦𝑟 (𝐼𝑉) = 16𝐴0 𝑒−2𝑡 𝑦(𝐼𝑉) − 6𝑦′′′ + 16𝑦′′ + 54𝑦′ − 225𝑦 = 100𝑒−2𝑡 16𝐴0 𝑒−2𝑡 + 48𝐴0 𝑒−2𝑡 + 64𝐴0 𝑒−2𝑡 − 108𝐴0 𝑒−2𝑡 − 225𝐴0 𝑒−2𝑡 = 100𝑒−2𝑡 −205𝐴0 𝑒−2𝑡 = 100𝑒−2𝑡 𝐴0 = − 100 205 = − 20 41 So, the value of yr is 𝑦𝑟 = − 20 41 𝑒−2𝑡 𝑦 = 𝑐1 𝑒3𝑡 + 𝑐2 𝑒−3𝑡 + 𝑒3𝑡 (𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) − 20 41 𝑒−2𝑡 Now, we are going to find the value of c1, c2, and c3. y(0) = 1 1 = 𝑐1 + 𝑐2 + 𝑐3 − 20 41 𝑐1 + 𝑐2 + 𝑐3 = 61 41 ....................................... (1) y’(0)=1 𝑦′ = 3𝑐1 𝑒3𝑡 − 3𝑐2 𝑒−3𝑡 + 3𝑒3𝑡(𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) + 𝑒3𝑡(−4𝑐3 sin 4𝑡 + 4𝑐4 cos 4𝑡) + 40 41 𝑒−2𝑡 1 = 3𝑐1 − 3𝑐2 + 3𝑐3 + 4𝑐4 + 40 41 3𝑐1 − 3𝑐2 + 3𝑐3 + 4𝑐4 = 1 41 ........................ (2) y’’(0)=1
- 13. 𝑦′′ = 9𝑐1 𝑒3𝑡 + 9𝑐2 𝑒−3𝑡 + 9𝑒3𝑡(𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) + 3𝑒3𝑡(−4𝑐3 sin 4𝑡 + 4𝑐4 cos 4𝑡) + 3𝑒3𝑡(−4𝑐3 sin 4𝑡 + 4𝑐4 cos 4𝑡) − 16𝑒3𝑡(𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) + 80 41 𝑒−2𝑡 1 = 9𝑐1 + 9𝑐2 + 9𝑐3 + 12𝑐4 + 12𝑐4 − 16𝑐3 − 80 41 1 = 9(𝑐1 + 𝑐2 + 𝑐3) + 24𝑐4 − 16𝑐3 − 80 41 121 41 = 9( 61 41 ) + 24𝑐4 − 16𝑐3 − 428 41 = 24𝑐4 − 16𝑐3 2𝑐3 − 3𝑐4 = 107 82 .......................................... (3) y’’’(0)=1 𝑦′′′ = 27𝑐1 𝑒3𝑡 − 27𝑐2 𝑒−3𝑡 + 75𝑒3𝑡(𝑐3 cos 4𝑡 + 𝑐4 sin 4𝑡) − 100𝑒3𝑡(−𝑐3 sin 4𝑡 + 𝑐4 cos 4𝑡) + 160 41 𝑒−2𝑡 1 = 27𝑐1 − 27𝑐2 + 75𝑐3 − 100𝑐4 + 160 41 27𝑐1 − 27𝑐2 + 75𝑐3 − 100𝑐4 = − 119 41 ........ (4) To get the value of c1, c2, c3, and c4, we can use elimination and substituion method from the above equations. From (4) and (2), we can get the new equation: 27𝑐1 − 27𝑐2 + 75𝑐3 − 100𝑐4 = − 119 41 ........ (4) 27𝑐1 − 27𝑐2 + 27𝑐3 + 36𝑐4 = 9 41 ................ (2) 43𝑐3 − 136𝑐4 = − 128 41 6𝑐3 − 176𝑐4 = − 16 41 ..................................... (5) From (5) and (3), we can get the value of c4. +
- 14. 6𝑐3 − 176𝑐4 = − 16 41 ..................................... (5) 6𝑐3 − 9𝑐4 = 321 82 ... ....................................... (3) −8𝑐4 = −32 − 321 82 𝑐4 = −353 656 From (3), we can get the value of c3. 2𝑐3 − 3( −353 656 ) = 107 82 .......................................... (3) 2𝑐3 = 1915 656 𝑐3 = 1915 1315 From (1), we get: 𝑐1 + 𝑐2 + 1915 1315 = 61 41 ....................................... (1) 𝑐1 + 𝑐2 = 37 1312 ................................................ (6) From (2), we get: 3𝑐1 − 3𝑐2 + 3( 1915 1315 ) + 4( −353 656 ) = 1 41 ........................ (2) 3𝑐1 − 3𝑐2 = 32 − 5745 − 2824 1312 3𝑐1 − 3𝑐2 = − 8537 1312 𝑐1 − 𝑐2 = − 8537 3936 ................................................ (7) From (6) and (7), we can get the value of c1 and c2 by elimination method. 𝑐1 + 𝑐2 = 37 1312 ................................................ (6) +
- 15. 𝑐1 − 𝑐2 = − 8537 3936 ............................................. (7) 2𝑐1 = −8537 + 111 3936 2𝑐1 = −8426 3936 𝑐1 = −4213 3936 From (7), we can get the value of c2. −4213 3936 − 𝑐2 = − 8537 3936 ............................................. (7) 𝑐2 = 8537 − 4213 3936 𝑐2 = 1081 656 After getting the value of c1, c2, c3, and c4, we get the characteristic value. 𝑦 = − 4213 3936 𝑒3𝑡 + 1081 656 𝑒−3𝑡 + 𝑒3𝑡 ( 1915 1315 cos 4𝑡 − 353 656 sin 4𝑡) − 20 41 𝑒−2𝑡 +
- 16. TASK IV (March 13th, 2014) 1. 𝑑2 𝑞 𝑑𝑡2 + 1000 𝑑𝑞 𝑑𝑡 + 25000𝑞 = 24 𝑞(0) = 𝑞′(0) = 0 2. 𝑑2 𝑦 𝑑𝑡2 − 4 𝑑𝑦 𝑑𝑡 + 𝑦 = 2𝑡3 + 3𝑡2 − 1 𝑦(0) = 𝑦′(0) = 1 Solution: 1. 𝑞′′ + 1000𝑞′ + 25000𝑞 = 24 Quadratic equation: 𝑡2 + 1000𝑡 + 25000 = 0 𝑡1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑡1,2 = −1000 ± √(1000)2 − 4(1)(25000) 2(1) 𝑡1,2 = −1000 ± √900000 2 𝑡1,2 = −1000 ± 300√10 2 𝑡1,2 = −500 ± 150√10 𝑡1 = −500 + 150√10 and 𝑡2 = −500 − 150√10 𝑞𝑙 = 𝑐1 𝑒(−500+150√10)𝑡 + 𝑐2 𝑒(−500−150√10)𝑡 𝑞 𝑟 = 𝐴0 𝑞′ 𝑟 = 0 𝑞′′ 𝑟 = 0 𝑞′′ + 1000𝑞′ + 25000𝑞 = 24 0 + 1000(0) + 25000(𝐴0) = 24 𝐴0 = 3 3125 , then 𝑞 𝑟 = 3 3125
- 17. 𝑞 = 𝑞𝑙 + 𝑞 𝑟 𝑞(𝑡) = 𝑐1 𝑒(−500+150√10)𝑡 + 𝑐2 𝑒(−500−150√10)𝑡 + 3 3125 𝑞(0) = 0 0 = 𝑐1 + 𝑐2 + 3 3125 𝑐1 + 𝑐2 = − 3 3125 ...................................(1) 𝑞′(𝑡) = (−500 + 150√10)𝑐1 𝑒(−500+150√10)𝑡 + (−500 − 150√10)𝑐2 𝑒(−500−150√10)𝑡 𝑞′(0) = 0 0 = (−500 + 150√10)𝑐1 + (−500 − 150√10)𝑐2 ........................... (2) By using elimination method, we can find the value of c1 from (1) and (2). (−500 − 150√10)𝑐1 + (−500 − 150√10)𝑐2 = − 3 3125 (−500 − 150√10) (−500 + 150√10)𝑐1 + (−500 − 150√10)𝑐2 = 0 −300√10𝑐1 = − 3 3125 (−500 − 150√10) 𝑐1 = −5√10 − 15 3125 By using (1), we can find the value of c2. 𝑐1 + 𝑐2 = − 3 3125 −5√10 − 15 3125 + 𝑐2 = − 3 3125 𝑐2 = 5√10 + 12 3125 After finding the value of c1 and c2, we get the equation. 𝑞(𝑡) = ( −5√10 − 15 3125 ) 𝑒(−500+150√10)𝑡 + ( 5√10 + 12 3125 ) 𝑒(−500−150√10)𝑡 + 3 3125 –
- 18. 2. 𝑦′′ − 4𝑦′ + 𝑦 = 2𝑡3 + 3𝑡2 − 1 Quadratic equation: 𝑡2 − 4𝑡 + 1 = 0 𝑡1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝑡1,2 = −(−4) ± √(−4)2 − 4(1)(1) 2(1) 𝑡1,2 = 4 ± √12 2 𝑡1,2 = 4 ± 2√3 2 𝑡1,2 = 2 ± √3 𝑡1 = 2 + √3 and 𝑡2 = 2 − √3 𝑦𝑙 = 𝑐1 𝑒(2+√3)𝑡 + 𝑐2 𝑒(2−√3)𝑡 𝑦𝑟 = 𝐴3 𝑡3 + 𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0 𝑦′ 𝑟 = 3𝐴3 𝑡2 + 2𝐴2 𝑡 + 𝐴1 𝑦′′ 𝑟 = 6𝐴3 𝑡 + 2𝐴2 𝑦′′ − 4𝑦′ + 𝑦 = 2𝑡3 + 3𝑡2 − 1 (6𝐴3 𝑡 + 2𝐴2) − 4(3𝐴3 𝑡2 + 2𝐴2 𝑡 + 𝐴1) + (𝐴3 𝑡3 + 𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0) = 2𝑡3 + 3𝑡2 − 1 𝐴3 𝑡3 + (−12𝐴3 + 𝐴2)𝑡2 + (6𝐴3 − 8𝐴2 + 𝐴1)𝑡 + 𝐴1 + 𝐴0 = 2𝑡3 + 3𝑡2 − 1 Equation similarity: 𝐴3 = 2 −12𝐴3 + 𝐴2 = 3 −12(2) + 𝐴2 = 3 𝐴2 = 27 6𝐴3 − 8𝐴2 + 𝐴1 = 0 6(2) − 8(27) + 𝐴1 = 0 𝐴1 = 204
- 19. 𝐴1 + 𝐴0 = −1 204 + 𝐴0 = −1 𝐴0 = −205 𝑦𝑟 = 2𝑡3 + 27𝑡2 + 204𝑡 − 205 𝑦 = 𝑦𝑙 + 𝑦𝑟 𝑦(𝑡) = 𝑐1 𝑒(2+√3)𝑡 + 𝑐2 𝑒(2−√3)𝑡 + 2𝑡3 + 27𝑡2 + 204𝑡 − 205 𝑦(0) = 1 1 = 𝑐1 + 𝑐2 − 205 𝑐1 + 𝑐2 = 206 .........................(1) 𝑦′(𝑡) = (2 + √3)𝑐1 𝑒(2+√3)𝑡 + (2 − √3)𝑐2 𝑒(2−√3)𝑡 + 6𝑡2 + 54𝑡 + 204 𝑦′(0) = 1 1 = (2 + √3)𝑐1 + (2 − √3)𝑐2 + 204 (2 + √3)𝑐1 + (2 − √3)𝑐2 = −203 ........................... (2) By using elimination and substitution method, the value of c1 and c2 can be obtained from (1) and (2). −203 = (2 + √3)𝑐1 + (2 − √3)𝑐2........................... (2) 206 = (2 − √3)𝑐1 + (2 − √3)𝑐2 ...................................(1) 2√3𝑐1 = −203 − 206(2 − √3) 2√3𝑐1 = −715 + 206√3 𝑐1 = −715√3 + 618 6 𝑐1 + 𝑐2 = 206 ( −715√3 + 618 6 ) + 𝑐2 = 206 𝑐2 = 715√3 + 618 6 𝑦(𝑡) = ( −715√3+618 6 ) 𝑒(2+√3)𝑡 + ( 715√3+618 6 ) 𝑒(2−√3)𝑡 + 2𝑡3 + 27𝑡2 + 204𝑡 − 205 –
- 20. TASK V (March 20th, 2014) 1. 𝑑2 𝑥 𝑑𝑡2 + 4 𝑑𝑥 𝑑𝑡 + 8𝑥 = (20𝑡2 + 16𝑡 − 78)𝑒2𝑡 y(0)=y’(0)=0 2. 𝑑3 𝑞 𝑑𝑡3 − 5 𝑑2 𝑞 𝑑𝑡2 + 25 𝑑𝑞 𝑑𝑡 − 125𝑞 = (−500𝑡2 + 465𝑡 − 387)𝑒2𝑡 q(0)=q’(0)= q’’(0)=0 Solution: 1. 𝑑2 𝑥 𝑑𝑡2 + 4 𝑑𝑥 𝑑𝑡 + 8𝑥 = (20𝑡2 + 16𝑡 − 78)𝑒2𝑡 y(0)=y’(0)=0 quadratic equation is 𝜆2 + 4𝜆 + 8 = 0 𝜆1,2 = −𝑏 ± √𝑏2 − 4𝑎𝑐 2𝑎 𝜆1,2 = −4 ± √42 − 4(1)(8) 2(1) 𝜆1,2 = −4 ± √−16 2 𝜆1,2 = −2 ± 2𝑖 𝑦𝑙 = 𝑒−2𝑡 (𝑐1 cos 2𝑡 + 𝑐2 sin 2𝑡) 𝑦𝑟 = (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)𝑒2𝑡 𝑦′ 𝑟 = (2𝐴2 𝑡 + 𝐴1)𝑒2𝑡 + (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)(2𝑒2𝑡) 𝑦′′ 𝑟 = 2𝐴2 𝑒2𝑡 + (2𝐴2 𝑡 + 𝐴1)(2𝑒2𝑡) + (2𝐴2 𝑡 + 𝐴1)(2𝑒2𝑡) + (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)(4𝑒2𝑡) 𝑦′′ + 4𝑦′ + 8𝑦 = (20𝑡2 + 16𝑡 − 78)𝑒2𝑡 2𝐴2 𝑒2𝑡 + (2𝐴2 𝑡 + 𝐴1)(2𝑒2𝑡) + (2𝐴2 𝑡 + 𝐴1)(2𝑒2𝑡) + (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)(4𝑒2𝑡) + 4{(2𝐴2 𝑡 + 𝐴1)𝑒2𝑡 + (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)(2𝑒2𝑡)} + 8{(𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)𝑒2𝑡} = (20𝑡2 + 16𝑡 − 78)𝑒2𝑡 (2𝐴2 + 8𝐴1 + 20𝐴0)𝑒2𝑡 + (16𝐴2 + 20𝐴1)𝑡𝑒2𝑡 + (20𝐴2)𝑡2 𝑒2𝑡 = (20𝑡2 + 16𝑡 − 78)𝑒2𝑡 20𝐴2 = 20 𝐴2 = 1
- 21. 16𝐴2 + 20𝐴1 = 16 16(1) + 20𝐴1 = 16 20𝐴1 = 0 𝐴1 = 0 2𝐴2 + 8𝐴1 + 20𝐴0 = −78 2(1) + 8(0) + 20𝐴0 = −78 20𝐴0 = −78 − 2 20𝐴0 = −80 𝐴0 = −4 𝑦𝑟 = (𝑡2 − 4)𝑒2𝑡 y = yl + yr 𝑦(𝑡) = 𝑒−2𝑡(𝑐1 cos 2𝑡 + 𝑐2 sin 2𝑡) + (𝑡2 − 4)𝑒2𝑡 𝑦′(𝑡) = (−2𝑒−2𝑡)(𝑐1 cos 2𝑡 + 𝑐2 sin 2𝑡) + 𝑒−2𝑡(−2𝑐1 sin 2𝑡 + 2𝑐2 cos 2𝑡) + 2𝑡𝑒2𝑡 + (𝑡2 − 4)(2𝑒2𝑡) 𝑦(0) = 0 0 = 𝑐1 − 4 𝑐1 = 4 𝑦′(0) = 0 0 = −2𝑐1 + 2𝑐2 − 8 8 = −2(4) + 2𝑐2 𝑐2 = 8 𝑦(𝑡) = 𝑒−2𝑡(4 cos 2𝑡 + 8 sin 2𝑡) + (𝑡2 − 4)𝑒2𝑡 2. 𝑑3 𝑞 𝑑𝑡3 − 5 𝑑2 𝑞 𝑑𝑡2 + 25 𝑑𝑞 𝑑𝑡 − 125𝑞 = (−500𝑡2 + 465𝑡 − 387)𝑒2𝑡 q(0)=q’(0)= q’’(0)=0 Quadratic equation is: 𝜆3 − 5𝜆2 + 25𝜆 − 125 = 0
- 22. 𝜆1 = 5 𝜆2 = 5𝑖 𝜆3 = −5𝑖 𝑦𝑙 = 𝑐1 𝑒5𝑡 + 𝑐2 cos 5𝑡 + 𝑐3 sin 5𝑡 𝑦𝑟 = (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)𝑒2𝑡 𝑦′ 𝑟 = (2𝐴2 𝑡 + 𝐴1)𝑒2𝑡 + (𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)(2𝑒2𝑡 ) = 2𝐴2 𝑡𝑒2𝑡 + 𝐴1 𝑒2𝑡 + 2𝐴2 𝑡2 𝑒2𝑡 + 2𝐴1 𝑡𝑒2𝑡 + 2𝐴0 𝑒2𝑡 𝑦′′ 𝑟 = 2𝐴2 𝑒2𝑡 + 8𝐴2 𝑡𝑒2𝑡 + 2𝐴1 𝑒2𝑡 + 4𝐴2 𝑡𝑒2𝑡 + 4𝐴2 𝑡2 𝑒2𝑡 + 2𝐴1 𝑒2𝑡 + 4𝐴1 𝑡𝑒2𝑡 + 4𝐴0 𝑒2𝑡 𝑦′′′ 𝑟 = 12𝐴2 𝑒2𝑡 + 12𝐴1 𝑒2𝑡 + 8𝐴0 𝑒2𝑡 + 24𝐴2 𝑡𝑒2𝑡 + 8𝐴1 𝑡𝑒2𝑡 + 8𝐴2 𝑡2 𝑒2𝑡 𝑦′′′ − 5𝑦′′ + 25𝑦′ − 125𝑦 = (−500𝑡2 + 465𝑡 − 387)𝑒2𝑡 12𝐴2 𝑒2𝑡 + 12𝐴1 𝑒2𝑡 + 8𝐴0 𝑒2𝑡 + 24𝐴2 𝑡𝑒2𝑡 + 8𝐴1 𝑡𝑒2𝑡 + 8𝐴2 𝑡2 𝑒2𝑡 − 5{2𝐴2 𝑒2𝑡 + 8𝐴2 𝑡𝑒2𝑡 + 2𝐴1 𝑒2𝑡 + 4𝐴2 𝑡𝑒2𝑡 + 4𝐴2 𝑡2 𝑒2𝑡 + 2𝐴1 𝑒2𝑡 + 4𝐴1 𝑡𝑒2𝑡 + 4𝐴0 𝑒2𝑡 } + 25{2𝐴2 𝑡𝑒2𝑡 + 𝐴1 𝑒2𝑡 + 2𝐴2 𝑡2 𝑒2𝑡 + 2𝐴1 𝑡𝑒2𝑡 + 2𝐴0 𝑒2𝑡} − 125{(𝐴2 𝑡2 + 𝐴1 𝑡 + 𝐴0)𝑒2𝑡} = (−500𝑡2 + 465𝑡 − 387)𝑒2𝑡 (2𝐴2 + 17𝐴1 − 87𝐴0)𝑒2𝑡 + (34𝐴2 − 87𝐴1)𝑡𝑒2𝑡 + (−87𝐴2)𝑡2 𝑒2𝑡 = −500𝑡2 𝑒2𝑡 + 465𝑡𝑒2𝑡 − 387𝑒2𝑡 −87𝐴2 = −500 𝐴2 = 500 87 34𝐴2 − 87𝐴1 = 465 34 ( 500 87 ) − 87𝐴1 = 465 𝐴1 = −23455 7569 2𝐴2 + 17𝐴1 − 87𝐴0 = −387 2 ( 500 87 ) + 17 ( −23455 7569 ) − 87𝐴0 = −387 𝐴0 = 2617468 658503
- 23. 𝑦𝑟 = ( 500 87 𝑡2 − 23455 7569 𝑡 + 2617468 658503 ) 𝑒2𝑡 𝑦 = 𝑦𝑙 + 𝑦𝑟 𝑦(𝑡) = 𝑐1 𝑒5𝑡 + 𝑐2 cos 5𝑡 + 𝑐3 sin 5𝑡 + ( 500 87 𝑡2 − 23455 7569 𝑡 + 2617468 658503 ) 𝑒2𝑡 𝑦′ (𝑡) = 5𝑐1 𝑒5𝑡 − 5𝑐2 sin 5𝑡 + 5𝑐3 cos 5𝑡 + ( 1000 87 𝑡 − 23455 7569 ) 𝑒2𝑡 + ( 500 87 𝑡2 − 23455 7569 𝑡 + 2617468 658503 ) (2𝑒2𝑡) 𝑦′′(𝑡) = 25𝑐1 𝑒5𝑡 − 25𝑐2 cos5𝑡 − 25𝑐3 sin5𝑡 + 1000 87 𝑒2𝑡 + ( 1000 87 𝑡 − 23455 7569 ) (2𝑒2𝑡) + ( 1000 87 𝑡 − 23455 7569 ) (2𝑒2𝑡) + ( 500 87 𝑡2 − 23455 7569 𝑡 + 2617468 658503 ) (4𝑒2𝑡) 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 + 2617468 658503 𝑐1 + 𝑐2 = − 2617468 658503 ................................... (1) 𝑦′(0) = 0 0 = 5𝑐1 + 5𝑐3 − 23455 7569 + 5234936 658503 5𝑐1 + 5𝑐3 = − 3194351 658503 𝑐1 + 𝑐3 = − 3194351 3292515 .................................... (2) 𝑦′′(0) = 0 0 = 25𝑐1 − 25𝑐2 + 1000 87 − 46910 7569 − 46910 7569 + 10469872 658503 25𝑐1 − 25𝑐2 = − 9876532 658503 .................................... (3) 25𝑐1 + 25𝑐2 = − 65436700 658503 ................................... (1) 50𝑐1 = − 75313232 658503 𝑐1 = − 75313232 32925150 +
- 24. From equation (1), we get the value of c2. 𝑐2 = − 2617468 658503 + 75313232 32925150 𝑐2 = − 55560168 32925150 From equation (2), we get the value of c3. 𝑐3 = − 3194351 3292515 + 75313232 32925150 𝑐3 = 43369722 32925150 𝑦(𝑡) = − 75313232 32925150 𝑒5𝑡 − 55560168 32925150 cos 5𝑡 + 43369722 32925150 sin 5𝑡 + ( 500 87 𝑡2 − 23455 7569 𝑡 + 2617468 658503 ) 𝑒2𝑡