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Task compilation - Differential Equation II

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Jazz Michele Pasaribu
Jazz Michele Pasaribu

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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DIFFERENTIAL EQUATION II TASK COMPILATION 3/27/2014 MARIA PRISCILLYA PASARIBU IDN. 4103312018
TASK I (February, 13th 2014) Determine the solution of:: 1. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 2 𝑑𝑦 𝑑𝑑 βˆ’ 5𝑦 = 0, y(0)=0; y’(0)=1 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 3 𝑑𝑦 𝑑𝑑 = 0, y(0)=0; y’(0)=1 3. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 βˆ’ 4𝑦 = 0, y(0)=0; y’(0)=1 Solution: 1. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 2 𝑑𝑦 𝑑𝑑 βˆ’ 5𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation: πœ†2 βˆ’ 2πœ† βˆ’ 5 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž = βˆ’(βˆ’2) Β± √(βˆ’2)2 βˆ’ 4(1)(βˆ’5) 2(1) = 2 Β± √4 + 20 2 = 2 Β± √24 2 = 2 Β± 2√6 2 = 1 Β± √6 πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = (1 + √6)𝑐1 𝑒(1+√6)π‘₯ + (1 βˆ’ √6)𝑐1 𝑒(1βˆ’βˆš6)π‘₯ 1 = (1 + √6)𝑐1 + (1 βˆ’ √6)𝑐2 1 = (1 + √6)𝑐1 + (1 βˆ’ √6)(βˆ’π‘1) 1 = 𝑐1(1 + √6 βˆ’ 1 + √6) 1 = 2√6𝑐1 𝑐1 = 1 12 √6
𝑐1 = βˆ’π‘2 .................... (1) 𝑐2 = βˆ’ 1 12 √6 Maka, 𝑦 = 1 12 √6𝑒(1+√6)π‘₯ βˆ’ 1 12 √6𝑒(1βˆ’βˆš6)π‘₯ 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 3 𝑑𝑦 𝑑𝑑 = 0, y(0)=0; y’(0)=1 Characteristic Equation:πœ†2 βˆ’ 3πœ† = 0 πœ†(πœ† βˆ’ 3) = 0 πœ†1 = 0 ∨ πœ†2 = 3 πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦 = 𝑐1 𝑒0π‘₯ + 𝑐2 𝑒3π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = 3𝑐2 3π‘₯ 1 = 3𝑐2 𝑐2 = 1 3 𝑐1 = βˆ’ 1 3 Then, 𝑦 = βˆ’ 1 3 𝑒 π‘š1 π‘₯ + 1 3 𝑒 π‘š2 π‘₯ 3. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 βˆ’ 4𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation:πœ†2 βˆ’ 4πœ† βˆ’ 4 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(βˆ’4) 2(1) = 4 Β± √16 + 16 2 = 4 Β± 4√2 2 = 2 Β± 2√2
πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦 = 𝑐1 𝑒(2+2√2)π‘₯ + 𝑐2 𝑒(2βˆ’2√2)π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = (2 + 2√2)𝑐1 𝑒(2+2√2)π‘₯ + (2 βˆ’ 2√2)𝑐2 𝑒(2βˆ’2√2)π‘₯ 1 = (2 + 2√2)𝑐1 + (2 βˆ’ 2√2)𝑐2 1 = (2 + 2√2)𝑐1 + (2 βˆ’ 2√2)(βˆ’π‘1) 1 = 𝑐1(2 + 2√2 βˆ’ 2 + 2√2) 1 = 𝑐1(4√2) 𝑐1 = 1 8 √2 𝑐1 = βˆ’π‘2 .................... (1) 𝑐2 = βˆ’ 1 8 √2 Maka, 𝑦 = 1 8 √2𝑒(2+2√2)π‘₯ βˆ’ 1 8 √2𝑒(2βˆ’2√2)π‘₯
TASK II (February, 20th 2014) Solve these equation: 1. 𝑑4 𝑦 𝑑π‘₯4 + 10 𝑑2 𝑦 𝑑π‘₯2 + 9𝑦 = 0 2. 𝑑4 𝑦 𝑑π‘₯4 + 𝑑3 𝑦 𝑑π‘₯3 + 𝑑2 𝑦 𝑑π‘₯2 + 2𝑦 = 0 3. 𝑦′′′ + 4𝑦′ = 0 4. 𝑦(4) + 4𝑦′′ βˆ’ 𝑦′ + 6𝑦 = 0 5. 𝑑6 𝑦 𝑑π‘₯6 βˆ’ 4 𝑑5 𝑦 𝑑π‘₯5 + 16 𝑑4 𝑦 𝑑π‘₯4 βˆ’ 12 𝑑3 𝑦 𝑑π‘₯3 + 41 𝑑2 𝑦 𝑑π‘₯2 βˆ’ 8 𝑑𝑦 𝑑π‘₯ + 26𝑦 = 0 Solution: 1. 𝑑4 𝑦 𝑑π‘₯4 + 10 𝑑2 𝑦 𝑑π‘₯2 + 9𝑦 = 0 Characteristic equation: πœ†4 + 10πœ†2 + 9 = 0 (πœ†2 + 9)(πœ†2 + 1) = 0 (πœ† + 3𝑖)(πœ† βˆ’ 3𝑖)(πœ† + 𝑖)(πœ† βˆ’ 𝑖) = 0 πœ†1 = βˆ’3𝑖 ⋁ πœ†2 = βˆ’3𝑖 ⋁ πœ†3 = βˆ’π‘– ⋁ πœ†3 = 𝑖 So, the solution is 𝑦 = 𝐢1 cos 3π‘₯ + 𝐢2 sin 3π‘₯ + 𝐢3 cos π‘₯ + 𝐢4 sin π‘₯ 2. 𝑑4 𝑦 𝑑π‘₯4 + 𝑑3 𝑦 𝑑π‘₯3 + 𝑑2 𝑦 𝑑π‘₯2 + 2𝑦 = 0 Characteristic equation: πœ†4 + πœ†3 + πœ†2 + 2 = 0 (πœ†2 βˆ’ πœ† + 1)(πœ†2 + 2πœ† + 2) = 0 ο‚· (πœ†2 βˆ’ πœ† + 1) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’(βˆ’1) Β± √(βˆ’1)2 βˆ’ 4(1)(1) 2(1) πœ†1,2 = 1 Β± √1 βˆ’ 4 2 πœ†1,2 = 1 Β± √3𝑖 2 πœ†1 = 1 2 + 1 2 √3𝑖 and πœ†2 = 1 2 βˆ’ 1 2 √3𝑖 ο‚· (πœ†2 + 2πœ† + 2) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž
πœ†3,4 = βˆ’(2) Β± √(2)2 βˆ’ 4(1)(2) 2(1) πœ†3,4 = βˆ’2 Β± √4 βˆ’ 8 2 πœ†3,4 = βˆ’2 Β± 2𝑖 2 πœ†3 = βˆ’1 + 𝑖 and πœ†4 = βˆ’1 βˆ’ 𝑖 So, the solution is 𝑦 = 𝑒 1 2 π‘₯ (𝐢1 cos 1 2 √3π‘₯ + 𝐢2 sin 1 2 √3π‘₯) + π‘’βˆ’π‘₯ (𝐢3 cos π‘₯ + 𝐢4 sin π‘₯) 3. 𝑦′′′ + 4𝑦′ = 0 Characteristic equation: πœ†3 + 4πœ† = 0 πœ†(πœ†2 + 4) = 0 πœ†(πœ† + 2𝑖)(πœ† βˆ’ 2𝑖) = 0 πœ†1 = 0 ⋁ πœ†2 = βˆ’2𝑖 ⋁ πœ†3 = 2𝑖 So, the solution is 𝑦 = 𝐢1 + 𝐢2cos2π‘₯ + 𝐢3 sin 2π‘₯ 4. 𝑦(4) + 4𝑦′′ βˆ’ 𝑦′ + 6𝑦 = 0 Characteristics equation is πœ†4 + 4πœ†2 βˆ’ πœ† + 6 = 0 (πœ†2 βˆ’ πœ† + 2)(πœ†2 + πœ† + 3) = 0 ο‚· (πœ†2 βˆ’ πœ† + 2) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’(βˆ’1) Β± √(βˆ’1)2 βˆ’ 4(1)(2) 2(1) πœ†1,2 = 1 Β± √1 βˆ’ 8 2 πœ†1,2 = 1 Β± √7𝑖 2 πœ†1 = 1 2 + 1 2 √7𝑖 and πœ†2 = 1 2 βˆ’ 1 2 √7𝑖 ο‚· (πœ†2 + πœ† + 3) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†3,4 = βˆ’(1) Β± √(1)2 βˆ’ 4(1)(3) 2(1)
πœ†3,4 = βˆ’1 Β± √1 βˆ’ 12 2 πœ†3,4 = βˆ’1 Β± √11𝑖 2 πœ†3 = βˆ’ 1 2 + 1 2 √11𝑖 and πœ†4 = βˆ’ 1 2 βˆ’ 1 2 √11𝑖 So, the equation is: 𝑦 = 𝑒 1 2 π‘₯ (𝐢1 cos 1 2 √7π‘₯ + 𝐢2 sin 1 2 √7π‘₯) + π‘’βˆ’ 1 2 π‘₯ (𝐢3 cos 1 2 √11 π‘₯ + 𝐢4 sin 1 2 √11 π‘₯) 5. 𝑑6 𝑦 𝑑π‘₯6 βˆ’ 4 𝑑5 𝑦 𝑑π‘₯5 + 16 𝑑4 𝑦 𝑑π‘₯4 βˆ’ 12 𝑑3 𝑦 𝑑π‘₯3 + 41 𝑑2 𝑦 𝑑π‘₯2 βˆ’ 8 𝑑𝑦 𝑑π‘₯ + 26𝑦 = 0 Characteristic equation is πœ†6 βˆ’ 4πœ†5 + 16πœ†4 βˆ’ 12πœ†3 + 41πœ†2 βˆ’ 8πœ† + 26 = 0 (πœ†4 + 3πœ†2 + 2)(πœ†2 βˆ’ 4πœ† + 13) = 0 (πœ†2 + 1)(πœ†2 + 2)(πœ†2 βˆ’ 4πœ† + 13) = 0 ο‚· (πœ†2 + 1) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = 0 Β± √0 βˆ’ 4(1)(1) 2(1) πœ†1,2 = Β±βˆšβˆ’4 2 πœ†1,2 = Β±2𝑖 2 πœ†1 = 𝑖 and πœ†2 = βˆ’π‘– ο‚· (πœ†2 + 2) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†3,4 = 0 Β± √0 βˆ’ 4(1)(2) 2(1)
πœ†3,4 = Β±βˆšβˆ’8 2 πœ†3,4 = Β±2√2𝑖 2 πœ†3 = √2𝑖 and πœ†4 = βˆ’βˆš2𝑖 ο‚· (πœ†2 βˆ’ 4πœ† + 13) = 0 πœ†5,6 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†5,6 = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(13) 2(1) πœ†5,6 = 4 Β± √16 βˆ’ 52 2 πœ†5,6 = 4 Β± 6𝑖 2 πœ†5 = 2 + 3𝑖 and πœ†6 = 2 βˆ’ 3𝑖 So, the solution is 𝑦 = 𝐢1 cos π‘₯ + 𝐢1 sin π‘₯ + 𝑒2π‘₯(𝐢3 cos 3π‘₯ + 𝐢4 sin 3π‘₯) + 𝐢5 cos √2π‘₯ + 𝐢5 sin √2π‘₯
TASK III (March 7th, 2014) 1. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = βˆ’60𝑒7𝑑 y(0)=0, y’(0)=1, y”(0)=2 2. 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2π‘₯ 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Solution: 1. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = βˆ’60𝑒7𝑑 y(0)=0, y’(0)=1, y”(0)=2 Quadratic equation: πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = βˆ’60𝑒7𝑑 Y= yl + yr πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = 0 (πœ† βˆ’ 5)(πœ† + 5𝑖)(πœ† βˆ’ 5𝑖) = 0 𝑦𝑙 = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 π‘¦π‘Ÿ = 𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€² = 7𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€²β€² = 49𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€²β€²β€² = 343𝐴0 𝑒7𝑑 𝑦′′′ βˆ’ 5𝑦′′ + 25𝑦 βˆ’ 125𝑦 = βˆ’60𝑒7𝑑 343𝐴0 𝑒7𝑑 βˆ’ 245𝐴0 𝑒7𝑑 + 175𝐴0 𝑒7𝑑 βˆ’ 125𝐴0 𝑒7𝑑 = βˆ’60𝑒7𝑑 148𝐴0 𝑒7𝑑 = βˆ’60𝑒7𝑑 𝐴0 = βˆ’ 60 148 = βˆ’ 15 37 π‘¦π‘Ÿ = βˆ’ 15 37 𝑒7𝑑
Then, we got the equation is equal to Y= yl + yr 𝑦(𝑑) = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 βˆ’ 15 37 𝑒7𝑑 Now, we are going to find the value of c1, c2, and c3. y(0)=0 0 = 𝑐1 + 𝑐2 βˆ’ 15 37 𝑐1 + 𝑐2 = 15 37 ........................................ (1) y’(0)=1 𝑦′ (𝑑) = 5𝑐1 𝑒5𝑑 βˆ’ 5𝑐2 sin 5𝑑 + 5𝑐3 cos 5𝑑 βˆ’ 105 37 𝑒7𝑑 1 = 5𝑐1 + 5𝑐3 βˆ’ 105 37 𝑐1 + 𝑐3 = 142 185 .................................... (2) y”(0)=2 𝑦′′ (𝑑) = 25𝑐1 𝑒5𝑑 βˆ’ 25𝑐2 cos 5𝑑 βˆ’ 25𝑐3 sin 5𝑑 βˆ’ 735 37 𝑒7𝑑 2 = 25𝑐1 βˆ’ 25𝑐2 βˆ’ 735 37 25𝑐1 βˆ’ 25𝑐2 = 809 37 𝑐1 βˆ’ 𝑐2 = 809 925 ....................................... (3) Elimination of 𝑐2 from (1) and (3) 𝑐1 βˆ’ 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 βˆ’ 𝑐2 = 809 925 ....................................... (3) +
592 925 βˆ’ 𝑐2 = 809 925 𝑐2 = βˆ’ 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(𝑑) = 592 925 𝑒5𝑑 βˆ’ 217 925 cos 5𝑑 + 118 925 sin 5𝑑 βˆ’ 15 37 𝑒7𝑑 2. 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2π‘₯ 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Quadratic equation: πœ†4 βˆ’ 6πœ†3 + 16πœ†2 + 54πœ† βˆ’ 225 = 100π‘’βˆ’2π‘₯ 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ πœ†4 βˆ’ 6πœ†3 + 16πœ†2 + 54πœ† βˆ’ 225 = 0 By using Horner method and ABC formula, so that the roots are gotten: πœ†1 = 3 πœ†3 = 3 + 4𝑖 πœ†2 = βˆ’3 πœ†4 = 3 βˆ’ 4𝑖 𝑦𝑙 = 𝑐1 𝑒3𝑑 + 𝑐2 π‘’βˆ’3𝑑 + 𝑒3𝑑 (𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) π‘¦π‘Ÿ = 𝐴0 π‘’βˆ’2π‘₯ π‘¦π‘Ÿ β€² = βˆ’2𝐴0 π‘’βˆ’2𝑑
π‘¦π‘Ÿ β€²β€² = 4𝐴0 π‘’βˆ’2𝑑 π‘¦π‘Ÿ β€²β€²β€² = βˆ’8𝐴0 π‘’βˆ’2𝑑 π‘¦π‘Ÿ (𝐼𝑉) = 16𝐴0 π‘’βˆ’2𝑑 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2𝑑 16𝐴0 π‘’βˆ’2𝑑 + 48𝐴0 π‘’βˆ’2𝑑 + 64𝐴0 π‘’βˆ’2𝑑 βˆ’ 108𝐴0 π‘’βˆ’2𝑑 βˆ’ 225𝐴0 π‘’βˆ’2𝑑 = 100π‘’βˆ’2𝑑 βˆ’205𝐴0 π‘’βˆ’2𝑑 = 100π‘’βˆ’2𝑑 𝐴0 = βˆ’ 100 205 = βˆ’ 20 41 So, the value of yr is π‘¦π‘Ÿ = βˆ’ 20 41 π‘’βˆ’2𝑑 𝑦 = 𝑐1 𝑒3𝑑 + 𝑐2 π‘’βˆ’3𝑑 + 𝑒3𝑑 (𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) βˆ’ 20 41 π‘’βˆ’2𝑑 Now, we are going to find the value of c1, c2, and c3. y(0) = 1 1 = 𝑐1 + 𝑐2 + 𝑐3 βˆ’ 20 41 𝑐1 + 𝑐2 + 𝑐3 = 61 41 ....................................... (1) y’(0)=1 𝑦′ = 3𝑐1 𝑒3𝑑 βˆ’ 3𝑐2 π‘’βˆ’3𝑑 + 3𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) + 40 41 π‘’βˆ’2𝑑 1 = 3𝑐1 βˆ’ 3𝑐2 + 3𝑐3 + 4𝑐4 + 40 41 3𝑐1 βˆ’ 3𝑐2 + 3𝑐3 + 4𝑐4 = 1 41 ........................ (2) y’’(0)=1
𝑦′′ = 9𝑐1 𝑒3𝑑 + 9𝑐2 π‘’βˆ’3𝑑 + 9𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 3𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) + 3𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) βˆ’ 16𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 80 41 π‘’βˆ’2𝑑 1 = 9𝑐1 + 9𝑐2 + 9𝑐3 + 12𝑐4 + 12𝑐4 βˆ’ 16𝑐3 βˆ’ 80 41 1 = 9(𝑐1 + 𝑐2 + 𝑐3) + 24𝑐4 βˆ’ 16𝑐3 βˆ’ 80 41 121 41 = 9( 61 41 ) + 24𝑐4 βˆ’ 16𝑐3 βˆ’ 428 41 = 24𝑐4 βˆ’ 16𝑐3 2𝑐3 βˆ’ 3𝑐4 = 107 82 .......................................... (3) y’’’(0)=1 𝑦′′′ = 27𝑐1 𝑒3𝑑 βˆ’ 27𝑐2 π‘’βˆ’3𝑑 + 75𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) βˆ’ 100𝑒3𝑑(βˆ’π‘3 sin 4𝑑 + 𝑐4 cos 4𝑑) + 160 41 π‘’βˆ’2𝑑 1 = 27𝑐1 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 + 160 41 27𝑐1 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 = βˆ’ 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 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 = βˆ’ 119 41 ........ (4) 27𝑐1 βˆ’ 27𝑐2 + 27𝑐3 + 36𝑐4 = 9 41 ................ (2) 43𝑐3 βˆ’ 136𝑐4 = βˆ’ 128 41 6𝑐3 βˆ’ 176𝑐4 = βˆ’ 16 41 ..................................... (5) From (5) and (3), we can get the value of c4. +
6𝑐3 βˆ’ 176𝑐4 = βˆ’ 16 41 ..................................... (5) 6𝑐3 βˆ’ 9𝑐4 = 321 82 ... ....................................... (3) βˆ’8𝑐4 = βˆ’32 βˆ’ 321 82 𝑐4 = βˆ’353 656 From (3), we can get the value of c3. 2𝑐3 βˆ’ 3( βˆ’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 βˆ’ 3𝑐2 + 3( 1915 1315 ) + 4( βˆ’353 656 ) = 1 41 ........................ (2) 3𝑐1 βˆ’ 3𝑐2 = 32 βˆ’ 5745 βˆ’ 2824 1312 3𝑐1 βˆ’ 3𝑐2 = βˆ’ 8537 1312 𝑐1 βˆ’ 𝑐2 = βˆ’ 8537 3936 ................................................ (7) From (6) and (7), we can get the value of c1 and c2 by elimination method. 𝑐1 + 𝑐2 = 37 1312 ................................................ (6) +
𝑐1 βˆ’ 𝑐2 = βˆ’ 8537 3936 ............................................. (7) 2𝑐1 = βˆ’8537 + 111 3936 2𝑐1 = βˆ’8426 3936 𝑐1 = βˆ’4213 3936 From (7), we can get the value of c2. βˆ’4213 3936 βˆ’ 𝑐2 = βˆ’ 8537 3936 ............................................. (7) 𝑐2 = 8537 βˆ’ 4213 3936 𝑐2 = 1081 656 After getting the value of c1, c2, c3, and c4, we get the characteristic value. 𝑦 = βˆ’ 4213 3936 𝑒3𝑑 + 1081 656 π‘’βˆ’3𝑑 + 𝑒3𝑑 ( 1915 1315 cos 4𝑑 βˆ’ 353 656 sin 4𝑑) βˆ’ 20 41 π‘’βˆ’2𝑑 +
TASK IV (March 13th, 2014) 1. 𝑑2 π‘ž 𝑑𝑑2 + 1000 π‘‘π‘ž 𝑑𝑑 + 25000π‘ž = 24 π‘ž(0) = π‘žβ€²(0) = 0 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 𝑦(0) = 𝑦′(0) = 1 Solution: 1. π‘žβ€²β€² + 1000π‘žβ€² + 25000π‘ž = 24 Quadratic equation: 𝑑2 + 1000𝑑 + 25000 = 0 𝑑1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž 𝑑1,2 = βˆ’1000 Β± √(1000)2 βˆ’ 4(1)(25000) 2(1) 𝑑1,2 = βˆ’1000 Β± √900000 2 𝑑1,2 = βˆ’1000 Β± 300√10 2 𝑑1,2 = βˆ’500 Β± 150√10 𝑑1 = βˆ’500 + 150√10 and 𝑑2 = βˆ’500 βˆ’ 150√10 π‘žπ‘™ = 𝑐1 𝑒(βˆ’500+150√10)𝑑 + 𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 π‘ž π‘Ÿ = 𝐴0 π‘žβ€² π‘Ÿ = 0 π‘žβ€²β€² π‘Ÿ = 0 π‘žβ€²β€² + 1000π‘žβ€² + 25000π‘ž = 24 0 + 1000(0) + 25000(𝐴0) = 24 𝐴0 = 3 3125 , then π‘ž π‘Ÿ = 3 3125
π‘ž = π‘žπ‘™ + π‘ž π‘Ÿ π‘ž(𝑑) = 𝑐1 𝑒(βˆ’500+150√10)𝑑 + 𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 + 3 3125 π‘ž(0) = 0 0 = 𝑐1 + 𝑐2 + 3 3125 𝑐1 + 𝑐2 = βˆ’ 3 3125 ...................................(1) π‘žβ€²(𝑑) = (βˆ’500 + 150√10)𝑐1 𝑒(βˆ’500+150√10)𝑑 + (βˆ’500 βˆ’ 150√10)𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 π‘žβ€²(0) = 0 0 = (βˆ’500 + 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 ........................... (2) By using elimination method, we can find the value of c1 from (1) and (2). (βˆ’500 βˆ’ 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 = βˆ’ 3 3125 (βˆ’500 βˆ’ 150√10) (βˆ’500 + 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 = 0 βˆ’300√10𝑐1 = βˆ’ 3 3125 (βˆ’500 βˆ’ 150√10) 𝑐1 = βˆ’5√10 βˆ’ 15 3125 By using (1), we can find the value of c2. 𝑐1 + 𝑐2 = βˆ’ 3 3125 βˆ’5√10 βˆ’ 15 3125 + 𝑐2 = βˆ’ 3 3125 𝑐2 = 5√10 + 12 3125 After finding the value of c1 and c2, we get the equation. π‘ž(𝑑) = ( βˆ’5√10 βˆ’ 15 3125 ) 𝑒(βˆ’500+150√10)𝑑 + ( 5√10 + 12 3125 ) 𝑒(βˆ’500βˆ’150√10)𝑑 + 3 3125 –
2. 𝑦′′ βˆ’ 4𝑦′ + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 Quadratic equation: 𝑑2 βˆ’ 4𝑑 + 1 = 0 𝑑1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž 𝑑1,2 = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(1) 2(1) 𝑑1,2 = 4 Β± √12 2 𝑑1,2 = 4 Β± 2√3 2 𝑑1,2 = 2 Β± √3 𝑑1 = 2 + √3 and 𝑑2 = 2 βˆ’ √3 𝑦𝑙 = 𝑐1 𝑒(2+√3)𝑑 + 𝑐2 𝑒(2βˆ’βˆš3)𝑑 π‘¦π‘Ÿ = 𝐴3 𝑑3 + 𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0 𝑦′ π‘Ÿ = 3𝐴3 𝑑2 + 2𝐴2 𝑑 + 𝐴1 𝑦′′ π‘Ÿ = 6𝐴3 𝑑 + 2𝐴2 𝑦′′ βˆ’ 4𝑦′ + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 (6𝐴3 𝑑 + 2𝐴2) βˆ’ 4(3𝐴3 𝑑2 + 2𝐴2 𝑑 + 𝐴1) + (𝐴3 𝑑3 + 𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0) = 2𝑑3 + 3𝑑2 βˆ’ 1 𝐴3 𝑑3 + (βˆ’12𝐴3 + 𝐴2)𝑑2 + (6𝐴3 βˆ’ 8𝐴2 + 𝐴1)𝑑 + 𝐴1 + 𝐴0 = 2𝑑3 + 3𝑑2 βˆ’ 1 Equation similarity: 𝐴3 = 2 βˆ’12𝐴3 + 𝐴2 = 3 βˆ’12(2) + 𝐴2 = 3 𝐴2 = 27 6𝐴3 βˆ’ 8𝐴2 + 𝐴1 = 0 6(2) βˆ’ 8(27) + 𝐴1 = 0 𝐴1 = 204
𝐴1 + 𝐴0 = βˆ’1 204 + 𝐴0 = βˆ’1 𝐴0 = βˆ’205 π‘¦π‘Ÿ = 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ 𝑦(𝑑) = 𝑐1 𝑒(2+√3)𝑑 + 𝑐2 𝑒(2βˆ’βˆš3)𝑑 + 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 𝑦(0) = 1 1 = 𝑐1 + 𝑐2 βˆ’ 205 𝑐1 + 𝑐2 = 206 .........................(1) 𝑦′(𝑑) = (2 + √3)𝑐1 𝑒(2+√3)𝑑 + (2 βˆ’ √3)𝑐2 𝑒(2βˆ’βˆš3)𝑑 + 6𝑑2 + 54𝑑 + 204 𝑦′(0) = 1 1 = (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2 + 204 (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2 = βˆ’203 ........................... (2) By using elimination and substitution method, the value of c1 and c2 can be obtained from (1) and (2). βˆ’203 = (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2........................... (2) 206 = (2 βˆ’ √3)𝑐1 + (2 βˆ’ √3)𝑐2 ...................................(1) 2√3𝑐1 = βˆ’203 βˆ’ 206(2 βˆ’ √3) 2√3𝑐1 = βˆ’715 + 206√3 𝑐1 = βˆ’715√3 + 618 6 𝑐1 + 𝑐2 = 206 ( βˆ’715√3 + 618 6 ) + 𝑐2 = 206 𝑐2 = 715√3 + 618 6 𝑦(𝑑) = ( βˆ’715√3+618 6 ) 𝑒(2+√3)𝑑 + ( 715√3+618 6 ) 𝑒(2βˆ’βˆš3)𝑑 + 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 –
TASK V (March 20th, 2014) 1. 𝑑2 π‘₯ 𝑑𝑑2 + 4 𝑑π‘₯ 𝑑𝑑 + 8π‘₯ = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 y(0)=y’(0)=0 2. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 q(0)=q’(0)= q’’(0)=0 Solution: 1. 𝑑2 π‘₯ 𝑑𝑑2 + 4 𝑑π‘₯ 𝑑𝑑 + 8π‘₯ = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 y(0)=y’(0)=0 quadratic equation is πœ†2 + 4πœ† + 8 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’4 Β± √42 βˆ’ 4(1)(8) 2(1) πœ†1,2 = βˆ’4 Β± βˆšβˆ’16 2 πœ†1,2 = βˆ’2 Β± 2𝑖 𝑦𝑙 = π‘’βˆ’2𝑑 (𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) π‘¦π‘Ÿ = (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑 𝑦′ π‘Ÿ = (2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑) 𝑦′′ π‘Ÿ = 2𝐴2 𝑒2𝑑 + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(4𝑒2𝑑) 𝑦′′ + 4𝑦′ + 8𝑦 = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 2𝐴2 𝑒2𝑑 + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(4𝑒2𝑑) + 4{(2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑)} + 8{(𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑} = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 (2𝐴2 + 8𝐴1 + 20𝐴0)𝑒2𝑑 + (16𝐴2 + 20𝐴1)𝑑𝑒2𝑑 + (20𝐴2)𝑑2 𝑒2𝑑 = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 ο‚· 20𝐴2 = 20 𝐴2 = 1
ο‚· 16𝐴2 + 20𝐴1 = 16 16(1) + 20𝐴1 = 16 20𝐴1 = 0 𝐴1 = 0 ο‚· 2𝐴2 + 8𝐴1 + 20𝐴0 = βˆ’78 2(1) + 8(0) + 20𝐴0 = βˆ’78 20𝐴0 = βˆ’78 βˆ’ 2 20𝐴0 = βˆ’80 𝐴0 = βˆ’4 π‘¦π‘Ÿ = (𝑑2 βˆ’ 4)𝑒2𝑑 y = yl + yr 𝑦(𝑑) = π‘’βˆ’2𝑑(𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) + (𝑑2 βˆ’ 4)𝑒2𝑑 𝑦′(𝑑) = (βˆ’2π‘’βˆ’2𝑑)(𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) + π‘’βˆ’2𝑑(βˆ’2𝑐1 sin 2𝑑 + 2𝑐2 cos 2𝑑) + 2𝑑𝑒2𝑑 + (𝑑2 βˆ’ 4)(2𝑒2𝑑) 𝑦(0) = 0 0 = 𝑐1 βˆ’ 4 𝑐1 = 4 𝑦′(0) = 0 0 = βˆ’2𝑐1 + 2𝑐2 βˆ’ 8 8 = βˆ’2(4) + 2𝑐2 𝑐2 = 8 𝑦(𝑑) = π‘’βˆ’2𝑑(4 cos 2𝑑 + 8 sin 2𝑑) + (𝑑2 βˆ’ 4)𝑒2𝑑 2. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 q(0)=q’(0)= q’’(0)=0 Quadratic equation is: πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = 0
πœ†1 = 5 πœ†2 = 5𝑖 πœ†3 = βˆ’5𝑖 𝑦𝑙 = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 π‘¦π‘Ÿ = (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑 𝑦′ π‘Ÿ = (2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑 ) = 2𝐴2 𝑑𝑒2𝑑 + 𝐴1 𝑒2𝑑 + 2𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑑𝑒2𝑑 + 2𝐴0 𝑒2𝑑 𝑦′′ π‘Ÿ = 2𝐴2 𝑒2𝑑 + 8𝐴2 𝑑𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴2 𝑑𝑒2𝑑 + 4𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴1 𝑑𝑒2𝑑 + 4𝐴0 𝑒2𝑑 𝑦′′′ π‘Ÿ = 12𝐴2 𝑒2𝑑 + 12𝐴1 𝑒2𝑑 + 8𝐴0 𝑒2𝑑 + 24𝐴2 𝑑𝑒2𝑑 + 8𝐴1 𝑑𝑒2𝑑 + 8𝐴2 𝑑2 𝑒2𝑑 𝑦′′′ βˆ’ 5𝑦′′ + 25𝑦′ βˆ’ 125𝑦 = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 12𝐴2 𝑒2𝑑 + 12𝐴1 𝑒2𝑑 + 8𝐴0 𝑒2𝑑 + 24𝐴2 𝑑𝑒2𝑑 + 8𝐴1 𝑑𝑒2𝑑 + 8𝐴2 𝑑2 𝑒2𝑑 βˆ’ 5{2𝐴2 𝑒2𝑑 + 8𝐴2 𝑑𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴2 𝑑𝑒2𝑑 + 4𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴1 𝑑𝑒2𝑑 + 4𝐴0 𝑒2𝑑 } + 25{2𝐴2 𝑑𝑒2𝑑 + 𝐴1 𝑒2𝑑 + 2𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑑𝑒2𝑑 + 2𝐴0 𝑒2𝑑} βˆ’ 125{(𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑} = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 (2𝐴2 + 17𝐴1 βˆ’ 87𝐴0)𝑒2𝑑 + (34𝐴2 βˆ’ 87𝐴1)𝑑𝑒2𝑑 + (βˆ’87𝐴2)𝑑2 𝑒2𝑑 = βˆ’500𝑑2 𝑒2𝑑 + 465𝑑𝑒2𝑑 βˆ’ 387𝑒2𝑑 ο‚· βˆ’87𝐴2 = βˆ’500 𝐴2 = 500 87 ο‚· 34𝐴2 βˆ’ 87𝐴1 = 465 34 ( 500 87 ) βˆ’ 87𝐴1 = 465 𝐴1 = βˆ’23455 7569 ο‚· 2𝐴2 + 17𝐴1 βˆ’ 87𝐴0 = βˆ’387 2 ( 500 87 ) + 17 ( βˆ’23455 7569 ) βˆ’ 87𝐴0 = βˆ’387 𝐴0 = 2617468 658503
π‘¦π‘Ÿ = ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ 𝑦(𝑑) = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑 𝑦′ (𝑑) = 5𝑐1 𝑒5𝑑 βˆ’ 5𝑐2 sin 5𝑑 + 5𝑐3 cos 5𝑑 + ( 1000 87 𝑑 βˆ’ 23455 7569 ) 𝑒2𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) (2𝑒2𝑑) 𝑦′′(𝑑) = 25𝑐1 𝑒5𝑑 βˆ’ 25𝑐2 cos5𝑑 βˆ’ 25𝑐3 sin5𝑑 + 1000 87 𝑒2𝑑 + ( 1000 87 𝑑 βˆ’ 23455 7569 ) (2𝑒2𝑑) + ( 1000 87 𝑑 βˆ’ 23455 7569 ) (2𝑒2𝑑) + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) (4𝑒2𝑑) 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 + 2617468 658503 𝑐1 + 𝑐2 = βˆ’ 2617468 658503 ................................... (1) 𝑦′(0) = 0 0 = 5𝑐1 + 5𝑐3 βˆ’ 23455 7569 + 5234936 658503 5𝑐1 + 5𝑐3 = βˆ’ 3194351 658503 𝑐1 + 𝑐3 = βˆ’ 3194351 3292515 .................................... (2) 𝑦′′(0) = 0 0 = 25𝑐1 βˆ’ 25𝑐2 + 1000 87 βˆ’ 46910 7569 βˆ’ 46910 7569 + 10469872 658503 25𝑐1 βˆ’ 25𝑐2 = βˆ’ 9876532 658503 .................................... (3) 25𝑐1 + 25𝑐2 = βˆ’ 65436700 658503 ................................... (1) 50𝑐1 = βˆ’ 75313232 658503 𝑐1 = βˆ’ 75313232 32925150 +
From equation (1), we get the value of c2. 𝑐2 = βˆ’ 2617468 658503 + 75313232 32925150 𝑐2 = βˆ’ 55560168 32925150 From equation (2), we get the value of c3. 𝑐3 = βˆ’ 3194351 3292515 + 75313232 32925150 𝑐3 = 43369722 32925150 𝑦(𝑑) = βˆ’ 75313232 32925150 𝑒5𝑑 βˆ’ 55560168 32925150 cos 5𝑑 + 43369722 32925150 sin 5𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑

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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 𝑑𝑑2 βˆ’ 2 𝑑𝑦 𝑑𝑑 βˆ’ 5𝑦 = 0, y(0)=0; y’(0)=1 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 3 𝑑𝑦 𝑑𝑑 = 0, y(0)=0; y’(0)=1 3. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 βˆ’ 4𝑦 = 0, y(0)=0; y’(0)=1 Solution: 1. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 2 𝑑𝑦 𝑑𝑑 βˆ’ 5𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation: πœ†2 βˆ’ 2πœ† βˆ’ 5 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž = βˆ’(βˆ’2) Β± √(βˆ’2)2 βˆ’ 4(1)(βˆ’5) 2(1) = 2 Β± √4 + 20 2 = 2 Β± √24 2 = 2 Β± 2√6 2 = 1 Β± √6 πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = (1 + √6)𝑐1 𝑒(1+√6)π‘₯ + (1 βˆ’ √6)𝑐1 𝑒(1βˆ’βˆš6)π‘₯ 1 = (1 + √6)𝑐1 + (1 βˆ’ √6)𝑐2 1 = (1 + √6)𝑐1 + (1 βˆ’ √6)(βˆ’π‘1) 1 = 𝑐1(1 + √6 βˆ’ 1 + √6) 1 = 2√6𝑐1 𝑐1 = 1 12 √6
  • 3. 𝑐1 = βˆ’π‘2 .................... (1) 𝑐2 = βˆ’ 1 12 √6 Maka, 𝑦 = 1 12 √6𝑒(1+√6)π‘₯ βˆ’ 1 12 √6𝑒(1βˆ’βˆš6)π‘₯ 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 3 𝑑𝑦 𝑑𝑑 = 0, y(0)=0; y’(0)=1 Characteristic Equation:πœ†2 βˆ’ 3πœ† = 0 πœ†(πœ† βˆ’ 3) = 0 πœ†1 = 0 ∨ πœ†2 = 3 πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦 = 𝑐1 𝑒0π‘₯ + 𝑐2 𝑒3π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = 3𝑐2 3π‘₯ 1 = 3𝑐2 𝑐2 = 1 3 𝑐1 = βˆ’ 1 3 Then, 𝑦 = βˆ’ 1 3 𝑒 π‘š1 π‘₯ + 1 3 𝑒 π‘š2 π‘₯ 3. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 βˆ’ 4𝑦 = 0, y(0)=0; y’(0)=1 Characteristic Equation:πœ†2 βˆ’ 4πœ† βˆ’ 4 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(βˆ’4) 2(1) = 4 Β± √16 + 16 2 = 4 Β± 4√2 2 = 2 Β± 2√2
  • 4. πœ†1and πœ†2 are real and distinct, then 𝑦 = 𝑐1 𝑒 π‘š1 π‘₯ + 𝑐2 𝑒 π‘š2 π‘₯ 𝑦 = 𝑐1 𝑒(2+2√2)π‘₯ + 𝑐2 𝑒(2βˆ’2√2)π‘₯ 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 𝑐1 = βˆ’π‘2 .................... (1) 𝑦′(0) = 1 𝑦′(π‘₯) = (2 + 2√2)𝑐1 𝑒(2+2√2)π‘₯ + (2 βˆ’ 2√2)𝑐2 𝑒(2βˆ’2√2)π‘₯ 1 = (2 + 2√2)𝑐1 + (2 βˆ’ 2√2)𝑐2 1 = (2 + 2√2)𝑐1 + (2 βˆ’ 2√2)(βˆ’π‘1) 1 = 𝑐1(2 + 2√2 βˆ’ 2 + 2√2) 1 = 𝑐1(4√2) 𝑐1 = 1 8 √2 𝑐1 = βˆ’π‘2 .................... (1) 𝑐2 = βˆ’ 1 8 √2 Maka, 𝑦 = 1 8 √2𝑒(2+2√2)π‘₯ βˆ’ 1 8 √2𝑒(2βˆ’2√2)π‘₯
  • 5. TASK II (February, 20th 2014) Solve these equation: 1. 𝑑4 𝑦 𝑑π‘₯4 + 10 𝑑2 𝑦 𝑑π‘₯2 + 9𝑦 = 0 2. 𝑑4 𝑦 𝑑π‘₯4 + 𝑑3 𝑦 𝑑π‘₯3 + 𝑑2 𝑦 𝑑π‘₯2 + 2𝑦 = 0 3. 𝑦′′′ + 4𝑦′ = 0 4. 𝑦(4) + 4𝑦′′ βˆ’ 𝑦′ + 6𝑦 = 0 5. 𝑑6 𝑦 𝑑π‘₯6 βˆ’ 4 𝑑5 𝑦 𝑑π‘₯5 + 16 𝑑4 𝑦 𝑑π‘₯4 βˆ’ 12 𝑑3 𝑦 𝑑π‘₯3 + 41 𝑑2 𝑦 𝑑π‘₯2 βˆ’ 8 𝑑𝑦 𝑑π‘₯ + 26𝑦 = 0 Solution: 1. 𝑑4 𝑦 𝑑π‘₯4 + 10 𝑑2 𝑦 𝑑π‘₯2 + 9𝑦 = 0 Characteristic equation: πœ†4 + 10πœ†2 + 9 = 0 (πœ†2 + 9)(πœ†2 + 1) = 0 (πœ† + 3𝑖)(πœ† βˆ’ 3𝑖)(πœ† + 𝑖)(πœ† βˆ’ 𝑖) = 0 πœ†1 = βˆ’3𝑖 ⋁ πœ†2 = βˆ’3𝑖 ⋁ πœ†3 = βˆ’π‘– ⋁ πœ†3 = 𝑖 So, the solution is 𝑦 = 𝐢1 cos 3π‘₯ + 𝐢2 sin 3π‘₯ + 𝐢3 cos π‘₯ + 𝐢4 sin π‘₯ 2. 𝑑4 𝑦 𝑑π‘₯4 + 𝑑3 𝑦 𝑑π‘₯3 + 𝑑2 𝑦 𝑑π‘₯2 + 2𝑦 = 0 Characteristic equation: πœ†4 + πœ†3 + πœ†2 + 2 = 0 (πœ†2 βˆ’ πœ† + 1)(πœ†2 + 2πœ† + 2) = 0 ο‚· (πœ†2 βˆ’ πœ† + 1) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’(βˆ’1) Β± √(βˆ’1)2 βˆ’ 4(1)(1) 2(1) πœ†1,2 = 1 Β± √1 βˆ’ 4 2 πœ†1,2 = 1 Β± √3𝑖 2 πœ†1 = 1 2 + 1 2 √3𝑖 and πœ†2 = 1 2 βˆ’ 1 2 √3𝑖 ο‚· (πœ†2 + 2πœ† + 2) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž
  • 6. πœ†3,4 = βˆ’(2) Β± √(2)2 βˆ’ 4(1)(2) 2(1) πœ†3,4 = βˆ’2 Β± √4 βˆ’ 8 2 πœ†3,4 = βˆ’2 Β± 2𝑖 2 πœ†3 = βˆ’1 + 𝑖 and πœ†4 = βˆ’1 βˆ’ 𝑖 So, the solution is 𝑦 = 𝑒 1 2 π‘₯ (𝐢1 cos 1 2 √3π‘₯ + 𝐢2 sin 1 2 √3π‘₯) + π‘’βˆ’π‘₯ (𝐢3 cos π‘₯ + 𝐢4 sin π‘₯) 3. 𝑦′′′ + 4𝑦′ = 0 Characteristic equation: πœ†3 + 4πœ† = 0 πœ†(πœ†2 + 4) = 0 πœ†(πœ† + 2𝑖)(πœ† βˆ’ 2𝑖) = 0 πœ†1 = 0 ⋁ πœ†2 = βˆ’2𝑖 ⋁ πœ†3 = 2𝑖 So, the solution is 𝑦 = 𝐢1 + 𝐢2cos2π‘₯ + 𝐢3 sin 2π‘₯ 4. 𝑦(4) + 4𝑦′′ βˆ’ 𝑦′ + 6𝑦 = 0 Characteristics equation is πœ†4 + 4πœ†2 βˆ’ πœ† + 6 = 0 (πœ†2 βˆ’ πœ† + 2)(πœ†2 + πœ† + 3) = 0 ο‚· (πœ†2 βˆ’ πœ† + 2) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’(βˆ’1) Β± √(βˆ’1)2 βˆ’ 4(1)(2) 2(1) πœ†1,2 = 1 Β± √1 βˆ’ 8 2 πœ†1,2 = 1 Β± √7𝑖 2 πœ†1 = 1 2 + 1 2 √7𝑖 and πœ†2 = 1 2 βˆ’ 1 2 √7𝑖 ο‚· (πœ†2 + πœ† + 3) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†3,4 = βˆ’(1) Β± √(1)2 βˆ’ 4(1)(3) 2(1)
  • 7. πœ†3,4 = βˆ’1 Β± √1 βˆ’ 12 2 πœ†3,4 = βˆ’1 Β± √11𝑖 2 πœ†3 = βˆ’ 1 2 + 1 2 √11𝑖 and πœ†4 = βˆ’ 1 2 βˆ’ 1 2 √11𝑖 So, the equation is: 𝑦 = 𝑒 1 2 π‘₯ (𝐢1 cos 1 2 √7π‘₯ + 𝐢2 sin 1 2 √7π‘₯) + π‘’βˆ’ 1 2 π‘₯ (𝐢3 cos 1 2 √11 π‘₯ + 𝐢4 sin 1 2 √11 π‘₯) 5. 𝑑6 𝑦 𝑑π‘₯6 βˆ’ 4 𝑑5 𝑦 𝑑π‘₯5 + 16 𝑑4 𝑦 𝑑π‘₯4 βˆ’ 12 𝑑3 𝑦 𝑑π‘₯3 + 41 𝑑2 𝑦 𝑑π‘₯2 βˆ’ 8 𝑑𝑦 𝑑π‘₯ + 26𝑦 = 0 Characteristic equation is πœ†6 βˆ’ 4πœ†5 + 16πœ†4 βˆ’ 12πœ†3 + 41πœ†2 βˆ’ 8πœ† + 26 = 0 (πœ†4 + 3πœ†2 + 2)(πœ†2 βˆ’ 4πœ† + 13) = 0 (πœ†2 + 1)(πœ†2 + 2)(πœ†2 βˆ’ 4πœ† + 13) = 0 ο‚· (πœ†2 + 1) = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = 0 Β± √0 βˆ’ 4(1)(1) 2(1) πœ†1,2 = Β±βˆšβˆ’4 2 πœ†1,2 = Β±2𝑖 2 πœ†1 = 𝑖 and πœ†2 = βˆ’π‘– ο‚· (πœ†2 + 2) = 0 πœ†3,4 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†3,4 = 0 Β± √0 βˆ’ 4(1)(2) 2(1)
  • 8. πœ†3,4 = Β±βˆšβˆ’8 2 πœ†3,4 = Β±2√2𝑖 2 πœ†3 = √2𝑖 and πœ†4 = βˆ’βˆš2𝑖 ο‚· (πœ†2 βˆ’ 4πœ† + 13) = 0 πœ†5,6 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†5,6 = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(13) 2(1) πœ†5,6 = 4 Β± √16 βˆ’ 52 2 πœ†5,6 = 4 Β± 6𝑖 2 πœ†5 = 2 + 3𝑖 and πœ†6 = 2 βˆ’ 3𝑖 So, the solution is 𝑦 = 𝐢1 cos π‘₯ + 𝐢1 sin π‘₯ + 𝑒2π‘₯(𝐢3 cos 3π‘₯ + 𝐢4 sin 3π‘₯) + 𝐢5 cos √2π‘₯ + 𝐢5 sin √2π‘₯
  • 9. TASK III (March 7th, 2014) 1. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = βˆ’60𝑒7𝑑 y(0)=0, y’(0)=1, y”(0)=2 2. 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2π‘₯ 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Solution: 1. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = βˆ’60𝑒7𝑑 y(0)=0, y’(0)=1, y”(0)=2 Quadratic equation: πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = βˆ’60𝑒7𝑑 Y= yl + yr πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = 0 (πœ† βˆ’ 5)(πœ† + 5𝑖)(πœ† βˆ’ 5𝑖) = 0 𝑦𝑙 = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 π‘¦π‘Ÿ = 𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€² = 7𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€²β€² = 49𝐴0 𝑒7𝑑 π‘¦π‘Ÿ β€²β€²β€² = 343𝐴0 𝑒7𝑑 𝑦′′′ βˆ’ 5𝑦′′ + 25𝑦 βˆ’ 125𝑦 = βˆ’60𝑒7𝑑 343𝐴0 𝑒7𝑑 βˆ’ 245𝐴0 𝑒7𝑑 + 175𝐴0 𝑒7𝑑 βˆ’ 125𝐴0 𝑒7𝑑 = βˆ’60𝑒7𝑑 148𝐴0 𝑒7𝑑 = βˆ’60𝑒7𝑑 𝐴0 = βˆ’ 60 148 = βˆ’ 15 37 π‘¦π‘Ÿ = βˆ’ 15 37 𝑒7𝑑
  • 10. Then, we got the equation is equal to Y= yl + yr 𝑦(𝑑) = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 βˆ’ 15 37 𝑒7𝑑 Now, we are going to find the value of c1, c2, and c3. y(0)=0 0 = 𝑐1 + 𝑐2 βˆ’ 15 37 𝑐1 + 𝑐2 = 15 37 ........................................ (1) y’(0)=1 𝑦′ (𝑑) = 5𝑐1 𝑒5𝑑 βˆ’ 5𝑐2 sin 5𝑑 + 5𝑐3 cos 5𝑑 βˆ’ 105 37 𝑒7𝑑 1 = 5𝑐1 + 5𝑐3 βˆ’ 105 37 𝑐1 + 𝑐3 = 142 185 .................................... (2) y”(0)=2 𝑦′′ (𝑑) = 25𝑐1 𝑒5𝑑 βˆ’ 25𝑐2 cos 5𝑑 βˆ’ 25𝑐3 sin 5𝑑 βˆ’ 735 37 𝑒7𝑑 2 = 25𝑐1 βˆ’ 25𝑐2 βˆ’ 735 37 25𝑐1 βˆ’ 25𝑐2 = 809 37 𝑐1 βˆ’ 𝑐2 = 809 925 ....................................... (3) Elimination of 𝑐2 from (1) and (3) 𝑐1 βˆ’ 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 βˆ’ 𝑐2 = 809 925 ....................................... (3) +
  • 11. 592 925 βˆ’ 𝑐2 = 809 925 𝑐2 = βˆ’ 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(𝑑) = 592 925 𝑒5𝑑 βˆ’ 217 925 cos 5𝑑 + 118 925 sin 5𝑑 βˆ’ 15 37 𝑒7𝑑 2. 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2π‘₯ 𝑦(0) = 𝑦′(0) = 𝑦′′(0) = 𝑦′′′(0) = 1 Quadratic equation: πœ†4 βˆ’ 6πœ†3 + 16πœ†2 + 54πœ† βˆ’ 225 = 100π‘’βˆ’2π‘₯ 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ πœ†4 βˆ’ 6πœ†3 + 16πœ†2 + 54πœ† βˆ’ 225 = 0 By using Horner method and ABC formula, so that the roots are gotten: πœ†1 = 3 πœ†3 = 3 + 4𝑖 πœ†2 = βˆ’3 πœ†4 = 3 βˆ’ 4𝑖 𝑦𝑙 = 𝑐1 𝑒3𝑑 + 𝑐2 π‘’βˆ’3𝑑 + 𝑒3𝑑 (𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) π‘¦π‘Ÿ = 𝐴0 π‘’βˆ’2π‘₯ π‘¦π‘Ÿ β€² = βˆ’2𝐴0 π‘’βˆ’2𝑑
  • 12. π‘¦π‘Ÿ β€²β€² = 4𝐴0 π‘’βˆ’2𝑑 π‘¦π‘Ÿ β€²β€²β€² = βˆ’8𝐴0 π‘’βˆ’2𝑑 π‘¦π‘Ÿ (𝐼𝑉) = 16𝐴0 π‘’βˆ’2𝑑 𝑦(𝐼𝑉) βˆ’ 6𝑦′′′ + 16𝑦′′ + 54𝑦′ βˆ’ 225𝑦 = 100π‘’βˆ’2𝑑 16𝐴0 π‘’βˆ’2𝑑 + 48𝐴0 π‘’βˆ’2𝑑 + 64𝐴0 π‘’βˆ’2𝑑 βˆ’ 108𝐴0 π‘’βˆ’2𝑑 βˆ’ 225𝐴0 π‘’βˆ’2𝑑 = 100π‘’βˆ’2𝑑 βˆ’205𝐴0 π‘’βˆ’2𝑑 = 100π‘’βˆ’2𝑑 𝐴0 = βˆ’ 100 205 = βˆ’ 20 41 So, the value of yr is π‘¦π‘Ÿ = βˆ’ 20 41 π‘’βˆ’2𝑑 𝑦 = 𝑐1 𝑒3𝑑 + 𝑐2 π‘’βˆ’3𝑑 + 𝑒3𝑑 (𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) βˆ’ 20 41 π‘’βˆ’2𝑑 Now, we are going to find the value of c1, c2, and c3. y(0) = 1 1 = 𝑐1 + 𝑐2 + 𝑐3 βˆ’ 20 41 𝑐1 + 𝑐2 + 𝑐3 = 61 41 ....................................... (1) y’(0)=1 𝑦′ = 3𝑐1 𝑒3𝑑 βˆ’ 3𝑐2 π‘’βˆ’3𝑑 + 3𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) + 40 41 π‘’βˆ’2𝑑 1 = 3𝑐1 βˆ’ 3𝑐2 + 3𝑐3 + 4𝑐4 + 40 41 3𝑐1 βˆ’ 3𝑐2 + 3𝑐3 + 4𝑐4 = 1 41 ........................ (2) y’’(0)=1
  • 13. 𝑦′′ = 9𝑐1 𝑒3𝑑 + 9𝑐2 π‘’βˆ’3𝑑 + 9𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 3𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) + 3𝑒3𝑑(βˆ’4𝑐3 sin 4𝑑 + 4𝑐4 cos 4𝑑) βˆ’ 16𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) + 80 41 π‘’βˆ’2𝑑 1 = 9𝑐1 + 9𝑐2 + 9𝑐3 + 12𝑐4 + 12𝑐4 βˆ’ 16𝑐3 βˆ’ 80 41 1 = 9(𝑐1 + 𝑐2 + 𝑐3) + 24𝑐4 βˆ’ 16𝑐3 βˆ’ 80 41 121 41 = 9( 61 41 ) + 24𝑐4 βˆ’ 16𝑐3 βˆ’ 428 41 = 24𝑐4 βˆ’ 16𝑐3 2𝑐3 βˆ’ 3𝑐4 = 107 82 .......................................... (3) y’’’(0)=1 𝑦′′′ = 27𝑐1 𝑒3𝑑 βˆ’ 27𝑐2 π‘’βˆ’3𝑑 + 75𝑒3𝑑(𝑐3 cos 4𝑑 + 𝑐4 sin 4𝑑) βˆ’ 100𝑒3𝑑(βˆ’π‘3 sin 4𝑑 + 𝑐4 cos 4𝑑) + 160 41 π‘’βˆ’2𝑑 1 = 27𝑐1 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 + 160 41 27𝑐1 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 = βˆ’ 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 βˆ’ 27𝑐2 + 75𝑐3 βˆ’ 100𝑐4 = βˆ’ 119 41 ........ (4) 27𝑐1 βˆ’ 27𝑐2 + 27𝑐3 + 36𝑐4 = 9 41 ................ (2) 43𝑐3 βˆ’ 136𝑐4 = βˆ’ 128 41 6𝑐3 βˆ’ 176𝑐4 = βˆ’ 16 41 ..................................... (5) From (5) and (3), we can get the value of c4. +
  • 14. 6𝑐3 βˆ’ 176𝑐4 = βˆ’ 16 41 ..................................... (5) 6𝑐3 βˆ’ 9𝑐4 = 321 82 ... ....................................... (3) βˆ’8𝑐4 = βˆ’32 βˆ’ 321 82 𝑐4 = βˆ’353 656 From (3), we can get the value of c3. 2𝑐3 βˆ’ 3( βˆ’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 βˆ’ 3𝑐2 + 3( 1915 1315 ) + 4( βˆ’353 656 ) = 1 41 ........................ (2) 3𝑐1 βˆ’ 3𝑐2 = 32 βˆ’ 5745 βˆ’ 2824 1312 3𝑐1 βˆ’ 3𝑐2 = βˆ’ 8537 1312 𝑐1 βˆ’ 𝑐2 = βˆ’ 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 βˆ’ 𝑐2 = βˆ’ 8537 3936 ............................................. (7) 2𝑐1 = βˆ’8537 + 111 3936 2𝑐1 = βˆ’8426 3936 𝑐1 = βˆ’4213 3936 From (7), we can get the value of c2. βˆ’4213 3936 βˆ’ 𝑐2 = βˆ’ 8537 3936 ............................................. (7) 𝑐2 = 8537 βˆ’ 4213 3936 𝑐2 = 1081 656 After getting the value of c1, c2, c3, and c4, we get the characteristic value. 𝑦 = βˆ’ 4213 3936 𝑒3𝑑 + 1081 656 π‘’βˆ’3𝑑 + 𝑒3𝑑 ( 1915 1315 cos 4𝑑 βˆ’ 353 656 sin 4𝑑) βˆ’ 20 41 π‘’βˆ’2𝑑 +
  • 16. TASK IV (March 13th, 2014) 1. 𝑑2 π‘ž 𝑑𝑑2 + 1000 π‘‘π‘ž 𝑑𝑑 + 25000π‘ž = 24 π‘ž(0) = π‘žβ€²(0) = 0 2. 𝑑2 𝑦 𝑑𝑑2 βˆ’ 4 𝑑𝑦 𝑑𝑑 + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 𝑦(0) = 𝑦′(0) = 1 Solution: 1. π‘žβ€²β€² + 1000π‘žβ€² + 25000π‘ž = 24 Quadratic equation: 𝑑2 + 1000𝑑 + 25000 = 0 𝑑1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž 𝑑1,2 = βˆ’1000 Β± √(1000)2 βˆ’ 4(1)(25000) 2(1) 𝑑1,2 = βˆ’1000 Β± √900000 2 𝑑1,2 = βˆ’1000 Β± 300√10 2 𝑑1,2 = βˆ’500 Β± 150√10 𝑑1 = βˆ’500 + 150√10 and 𝑑2 = βˆ’500 βˆ’ 150√10 π‘žπ‘™ = 𝑐1 𝑒(βˆ’500+150√10)𝑑 + 𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 π‘ž π‘Ÿ = 𝐴0 π‘žβ€² π‘Ÿ = 0 π‘žβ€²β€² π‘Ÿ = 0 π‘žβ€²β€² + 1000π‘žβ€² + 25000π‘ž = 24 0 + 1000(0) + 25000(𝐴0) = 24 𝐴0 = 3 3125 , then π‘ž π‘Ÿ = 3 3125
  • 17. π‘ž = π‘žπ‘™ + π‘ž π‘Ÿ π‘ž(𝑑) = 𝑐1 𝑒(βˆ’500+150√10)𝑑 + 𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 + 3 3125 π‘ž(0) = 0 0 = 𝑐1 + 𝑐2 + 3 3125 𝑐1 + 𝑐2 = βˆ’ 3 3125 ...................................(1) π‘žβ€²(𝑑) = (βˆ’500 + 150√10)𝑐1 𝑒(βˆ’500+150√10)𝑑 + (βˆ’500 βˆ’ 150√10)𝑐2 𝑒(βˆ’500βˆ’150√10)𝑑 π‘žβ€²(0) = 0 0 = (βˆ’500 + 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 ........................... (2) By using elimination method, we can find the value of c1 from (1) and (2). (βˆ’500 βˆ’ 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 = βˆ’ 3 3125 (βˆ’500 βˆ’ 150√10) (βˆ’500 + 150√10)𝑐1 + (βˆ’500 βˆ’ 150√10)𝑐2 = 0 βˆ’300√10𝑐1 = βˆ’ 3 3125 (βˆ’500 βˆ’ 150√10) 𝑐1 = βˆ’5√10 βˆ’ 15 3125 By using (1), we can find the value of c2. 𝑐1 + 𝑐2 = βˆ’ 3 3125 βˆ’5√10 βˆ’ 15 3125 + 𝑐2 = βˆ’ 3 3125 𝑐2 = 5√10 + 12 3125 After finding the value of c1 and c2, we get the equation. π‘ž(𝑑) = ( βˆ’5√10 βˆ’ 15 3125 ) 𝑒(βˆ’500+150√10)𝑑 + ( 5√10 + 12 3125 ) 𝑒(βˆ’500βˆ’150√10)𝑑 + 3 3125 –
  • 18. 2. 𝑦′′ βˆ’ 4𝑦′ + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 Quadratic equation: 𝑑2 βˆ’ 4𝑑 + 1 = 0 𝑑1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž 𝑑1,2 = βˆ’(βˆ’4) Β± √(βˆ’4)2 βˆ’ 4(1)(1) 2(1) 𝑑1,2 = 4 Β± √12 2 𝑑1,2 = 4 Β± 2√3 2 𝑑1,2 = 2 Β± √3 𝑑1 = 2 + √3 and 𝑑2 = 2 βˆ’ √3 𝑦𝑙 = 𝑐1 𝑒(2+√3)𝑑 + 𝑐2 𝑒(2βˆ’βˆš3)𝑑 π‘¦π‘Ÿ = 𝐴3 𝑑3 + 𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0 𝑦′ π‘Ÿ = 3𝐴3 𝑑2 + 2𝐴2 𝑑 + 𝐴1 𝑦′′ π‘Ÿ = 6𝐴3 𝑑 + 2𝐴2 𝑦′′ βˆ’ 4𝑦′ + 𝑦 = 2𝑑3 + 3𝑑2 βˆ’ 1 (6𝐴3 𝑑 + 2𝐴2) βˆ’ 4(3𝐴3 𝑑2 + 2𝐴2 𝑑 + 𝐴1) + (𝐴3 𝑑3 + 𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0) = 2𝑑3 + 3𝑑2 βˆ’ 1 𝐴3 𝑑3 + (βˆ’12𝐴3 + 𝐴2)𝑑2 + (6𝐴3 βˆ’ 8𝐴2 + 𝐴1)𝑑 + 𝐴1 + 𝐴0 = 2𝑑3 + 3𝑑2 βˆ’ 1 Equation similarity: 𝐴3 = 2 βˆ’12𝐴3 + 𝐴2 = 3 βˆ’12(2) + 𝐴2 = 3 𝐴2 = 27 6𝐴3 βˆ’ 8𝐴2 + 𝐴1 = 0 6(2) βˆ’ 8(27) + 𝐴1 = 0 𝐴1 = 204
  • 19. 𝐴1 + 𝐴0 = βˆ’1 204 + 𝐴0 = βˆ’1 𝐴0 = βˆ’205 π‘¦π‘Ÿ = 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ 𝑦(𝑑) = 𝑐1 𝑒(2+√3)𝑑 + 𝑐2 𝑒(2βˆ’βˆš3)𝑑 + 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 𝑦(0) = 1 1 = 𝑐1 + 𝑐2 βˆ’ 205 𝑐1 + 𝑐2 = 206 .........................(1) 𝑦′(𝑑) = (2 + √3)𝑐1 𝑒(2+√3)𝑑 + (2 βˆ’ √3)𝑐2 𝑒(2βˆ’βˆš3)𝑑 + 6𝑑2 + 54𝑑 + 204 𝑦′(0) = 1 1 = (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2 + 204 (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2 = βˆ’203 ........................... (2) By using elimination and substitution method, the value of c1 and c2 can be obtained from (1) and (2). βˆ’203 = (2 + √3)𝑐1 + (2 βˆ’ √3)𝑐2........................... (2) 206 = (2 βˆ’ √3)𝑐1 + (2 βˆ’ √3)𝑐2 ...................................(1) 2√3𝑐1 = βˆ’203 βˆ’ 206(2 βˆ’ √3) 2√3𝑐1 = βˆ’715 + 206√3 𝑐1 = βˆ’715√3 + 618 6 𝑐1 + 𝑐2 = 206 ( βˆ’715√3 + 618 6 ) + 𝑐2 = 206 𝑐2 = 715√3 + 618 6 𝑦(𝑑) = ( βˆ’715√3+618 6 ) 𝑒(2+√3)𝑑 + ( 715√3+618 6 ) 𝑒(2βˆ’βˆš3)𝑑 + 2𝑑3 + 27𝑑2 + 204𝑑 βˆ’ 205 –
  • 20. TASK V (March 20th, 2014) 1. 𝑑2 π‘₯ 𝑑𝑑2 + 4 𝑑π‘₯ 𝑑𝑑 + 8π‘₯ = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 y(0)=y’(0)=0 2. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 q(0)=q’(0)= q’’(0)=0 Solution: 1. 𝑑2 π‘₯ 𝑑𝑑2 + 4 𝑑π‘₯ 𝑑𝑑 + 8π‘₯ = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 y(0)=y’(0)=0 quadratic equation is πœ†2 + 4πœ† + 8 = 0 πœ†1,2 = βˆ’π‘ Β± βˆšπ‘2 βˆ’ 4π‘Žπ‘ 2π‘Ž πœ†1,2 = βˆ’4 Β± √42 βˆ’ 4(1)(8) 2(1) πœ†1,2 = βˆ’4 Β± βˆšβˆ’16 2 πœ†1,2 = βˆ’2 Β± 2𝑖 𝑦𝑙 = π‘’βˆ’2𝑑 (𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) π‘¦π‘Ÿ = (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑 𝑦′ π‘Ÿ = (2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑) 𝑦′′ π‘Ÿ = 2𝐴2 𝑒2𝑑 + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(4𝑒2𝑑) 𝑦′′ + 4𝑦′ + 8𝑦 = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 2𝐴2 𝑒2𝑑 + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (2𝐴2 𝑑 + 𝐴1)(2𝑒2𝑑) + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(4𝑒2𝑑) + 4{(2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑)} + 8{(𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑} = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 (2𝐴2 + 8𝐴1 + 20𝐴0)𝑒2𝑑 + (16𝐴2 + 20𝐴1)𝑑𝑒2𝑑 + (20𝐴2)𝑑2 𝑒2𝑑 = (20𝑑2 + 16𝑑 βˆ’ 78)𝑒2𝑑 ο‚· 20𝐴2 = 20 𝐴2 = 1
  • 21. ο‚· 16𝐴2 + 20𝐴1 = 16 16(1) + 20𝐴1 = 16 20𝐴1 = 0 𝐴1 = 0 ο‚· 2𝐴2 + 8𝐴1 + 20𝐴0 = βˆ’78 2(1) + 8(0) + 20𝐴0 = βˆ’78 20𝐴0 = βˆ’78 βˆ’ 2 20𝐴0 = βˆ’80 𝐴0 = βˆ’4 π‘¦π‘Ÿ = (𝑑2 βˆ’ 4)𝑒2𝑑 y = yl + yr 𝑦(𝑑) = π‘’βˆ’2𝑑(𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) + (𝑑2 βˆ’ 4)𝑒2𝑑 𝑦′(𝑑) = (βˆ’2π‘’βˆ’2𝑑)(𝑐1 cos 2𝑑 + 𝑐2 sin 2𝑑) + π‘’βˆ’2𝑑(βˆ’2𝑐1 sin 2𝑑 + 2𝑐2 cos 2𝑑) + 2𝑑𝑒2𝑑 + (𝑑2 βˆ’ 4)(2𝑒2𝑑) 𝑦(0) = 0 0 = 𝑐1 βˆ’ 4 𝑐1 = 4 𝑦′(0) = 0 0 = βˆ’2𝑐1 + 2𝑐2 βˆ’ 8 8 = βˆ’2(4) + 2𝑐2 𝑐2 = 8 𝑦(𝑑) = π‘’βˆ’2𝑑(4 cos 2𝑑 + 8 sin 2𝑑) + (𝑑2 βˆ’ 4)𝑒2𝑑 2. 𝑑3 π‘ž 𝑑𝑑3 βˆ’ 5 𝑑2 π‘ž 𝑑𝑑2 + 25 π‘‘π‘ž 𝑑𝑑 βˆ’ 125π‘ž = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 q(0)=q’(0)= q’’(0)=0 Quadratic equation is: πœ†3 βˆ’ 5πœ†2 + 25πœ† βˆ’ 125 = 0
  • 22. πœ†1 = 5 πœ†2 = 5𝑖 πœ†3 = βˆ’5𝑖 𝑦𝑙 = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 π‘¦π‘Ÿ = (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑 𝑦′ π‘Ÿ = (2𝐴2 𝑑 + 𝐴1)𝑒2𝑑 + (𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)(2𝑒2𝑑 ) = 2𝐴2 𝑑𝑒2𝑑 + 𝐴1 𝑒2𝑑 + 2𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑑𝑒2𝑑 + 2𝐴0 𝑒2𝑑 𝑦′′ π‘Ÿ = 2𝐴2 𝑒2𝑑 + 8𝐴2 𝑑𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴2 𝑑𝑒2𝑑 + 4𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴1 𝑑𝑒2𝑑 + 4𝐴0 𝑒2𝑑 𝑦′′′ π‘Ÿ = 12𝐴2 𝑒2𝑑 + 12𝐴1 𝑒2𝑑 + 8𝐴0 𝑒2𝑑 + 24𝐴2 𝑑𝑒2𝑑 + 8𝐴1 𝑑𝑒2𝑑 + 8𝐴2 𝑑2 𝑒2𝑑 𝑦′′′ βˆ’ 5𝑦′′ + 25𝑦′ βˆ’ 125𝑦 = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 12𝐴2 𝑒2𝑑 + 12𝐴1 𝑒2𝑑 + 8𝐴0 𝑒2𝑑 + 24𝐴2 𝑑𝑒2𝑑 + 8𝐴1 𝑑𝑒2𝑑 + 8𝐴2 𝑑2 𝑒2𝑑 βˆ’ 5{2𝐴2 𝑒2𝑑 + 8𝐴2 𝑑𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴2 𝑑𝑒2𝑑 + 4𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑒2𝑑 + 4𝐴1 𝑑𝑒2𝑑 + 4𝐴0 𝑒2𝑑 } + 25{2𝐴2 𝑑𝑒2𝑑 + 𝐴1 𝑒2𝑑 + 2𝐴2 𝑑2 𝑒2𝑑 + 2𝐴1 𝑑𝑒2𝑑 + 2𝐴0 𝑒2𝑑} βˆ’ 125{(𝐴2 𝑑2 + 𝐴1 𝑑 + 𝐴0)𝑒2𝑑} = (βˆ’500𝑑2 + 465𝑑 βˆ’ 387)𝑒2𝑑 (2𝐴2 + 17𝐴1 βˆ’ 87𝐴0)𝑒2𝑑 + (34𝐴2 βˆ’ 87𝐴1)𝑑𝑒2𝑑 + (βˆ’87𝐴2)𝑑2 𝑒2𝑑 = βˆ’500𝑑2 𝑒2𝑑 + 465𝑑𝑒2𝑑 βˆ’ 387𝑒2𝑑 ο‚· βˆ’87𝐴2 = βˆ’500 𝐴2 = 500 87 ο‚· 34𝐴2 βˆ’ 87𝐴1 = 465 34 ( 500 87 ) βˆ’ 87𝐴1 = 465 𝐴1 = βˆ’23455 7569 ο‚· 2𝐴2 + 17𝐴1 βˆ’ 87𝐴0 = βˆ’387 2 ( 500 87 ) + 17 ( βˆ’23455 7569 ) βˆ’ 87𝐴0 = βˆ’387 𝐴0 = 2617468 658503
  • 23. π‘¦π‘Ÿ = ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑 𝑦 = 𝑦𝑙 + π‘¦π‘Ÿ 𝑦(𝑑) = 𝑐1 𝑒5𝑑 + 𝑐2 cos 5𝑑 + 𝑐3 sin 5𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑 𝑦′ (𝑑) = 5𝑐1 𝑒5𝑑 βˆ’ 5𝑐2 sin 5𝑑 + 5𝑐3 cos 5𝑑 + ( 1000 87 𝑑 βˆ’ 23455 7569 ) 𝑒2𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) (2𝑒2𝑑) 𝑦′′(𝑑) = 25𝑐1 𝑒5𝑑 βˆ’ 25𝑐2 cos5𝑑 βˆ’ 25𝑐3 sin5𝑑 + 1000 87 𝑒2𝑑 + ( 1000 87 𝑑 βˆ’ 23455 7569 ) (2𝑒2𝑑) + ( 1000 87 𝑑 βˆ’ 23455 7569 ) (2𝑒2𝑑) + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) (4𝑒2𝑑) 𝑦(0) = 0 0 = 𝑐1 + 𝑐2 + 2617468 658503 𝑐1 + 𝑐2 = βˆ’ 2617468 658503 ................................... (1) 𝑦′(0) = 0 0 = 5𝑐1 + 5𝑐3 βˆ’ 23455 7569 + 5234936 658503 5𝑐1 + 5𝑐3 = βˆ’ 3194351 658503 𝑐1 + 𝑐3 = βˆ’ 3194351 3292515 .................................... (2) 𝑦′′(0) = 0 0 = 25𝑐1 βˆ’ 25𝑐2 + 1000 87 βˆ’ 46910 7569 βˆ’ 46910 7569 + 10469872 658503 25𝑐1 βˆ’ 25𝑐2 = βˆ’ 9876532 658503 .................................... (3) 25𝑐1 + 25𝑐2 = βˆ’ 65436700 658503 ................................... (1) 50𝑐1 = βˆ’ 75313232 658503 𝑐1 = βˆ’ 75313232 32925150 +
  • 24. From equation (1), we get the value of c2. 𝑐2 = βˆ’ 2617468 658503 + 75313232 32925150 𝑐2 = βˆ’ 55560168 32925150 From equation (2), we get the value of c3. 𝑐3 = βˆ’ 3194351 3292515 + 75313232 32925150 𝑐3 = 43369722 32925150 𝑦(𝑑) = βˆ’ 75313232 32925150 𝑒5𝑑 βˆ’ 55560168 32925150 cos 5𝑑 + 43369722 32925150 sin 5𝑑 + ( 500 87 𝑑2 βˆ’ 23455 7569 𝑑 + 2617468 658503 ) 𝑒2𝑑