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# Task compilation - Differential Equation II

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Task compilation of differential equation II lecture

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### Task compilation - Differential Equation II

1. 1. DIFFERENTIAL EQUATION II TASK COMPILATION 3/27/2014 MARIA PRISCILLYA PASARIBU IDN. 4103312018
2. 2. TASK I (February, 13th 2014) Determine the solution of:: 1. π2 π¦ ππ‘2 β 2 ππ¦ ππ‘ β 5π¦ = 0, y(0)=0; yβ(0)=1 2. π2 π¦ ππ‘2 β 3 ππ¦ ππ‘ = 0, y(0)=0; yβ(0)=1 3. π2 π¦ ππ‘2 β 4 ππ¦ ππ‘ β 4π¦ = 0, y(0)=0; yβ(0)=1 Solution: 1. π2 π¦ ππ‘2 β 2 ππ¦ ππ‘ β 5π¦ = 0, y(0)=0; yβ(0)=1 Characteristic Equation: π2 β 2π β 5 = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π = β(β2) Β± β(β2)2 β 4(1)(β5) 2(1) = 2 Β± β4 + 20 2 = 2 Β± β24 2 = 2 Β± 2β6 2 = 1 Β± β6 π1and π2 are real and distinct, then π¦ = π1 π π1 π₯ + π2 π π2 π₯ π¦(0) = 0 0 = π1 + π2 π1 = βπ2 .................... (1) π¦β²(0) = 1 π¦β²(π₯) = (1 + β6)π1 π(1+β6)π₯ + (1 β β6)π1 π(1ββ6)π₯ 1 = (1 + β6)π1 + (1 β β6)π2 1 = (1 + β6)π1 + (1 β β6)(βπ1) 1 = π1(1 + β6 β 1 + β6) 1 = 2β6π1 π1 = 1 12 β6
3. 3. π1 = βπ2 .................... (1) π2 = β 1 12 β6 Maka, π¦ = 1 12 β6π(1+β6)π₯ β 1 12 β6π(1ββ6)π₯ 2. π2 π¦ ππ‘2 β 3 ππ¦ ππ‘ = 0, y(0)=0; yβ(0)=1 Characteristic Equation:π2 β 3π = 0 π(π β 3) = 0 π1 = 0 β¨ π2 = 3 π1and π2 are real and distinct, then π¦ = π1 π π1 π₯ + π2 π π2 π₯ π¦ = π1 π0π₯ + π2 π3π₯ π¦(0) = 0 0 = π1 + π2 π1 = βπ2 .................... (1) π¦β²(0) = 1 π¦β²(π₯) = 3π2 3π₯ 1 = 3π2 π2 = 1 3 π1 = β 1 3 Then, π¦ = β 1 3 π π1 π₯ + 1 3 π π2 π₯ 3. π2 π¦ ππ‘2 β 4 ππ¦ ππ‘ β 4π¦ = 0, y(0)=0; yβ(0)=1 Characteristic Equation:π2 β 4π β 4 = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π = β(β4) Β± β(β4)2 β 4(1)(β4) 2(1) = 4 Β± β16 + 16 2 = 4 Β± 4β2 2 = 2 Β± 2β2
4. 4. π1and π2 are real and distinct, then π¦ = π1 π π1 π₯ + π2 π π2 π₯ π¦ = π1 π(2+2β2)π₯ + π2 π(2β2β2)π₯ π¦(0) = 0 0 = π1 + π2 π1 = βπ2 .................... (1) π¦β²(0) = 1 π¦β²(π₯) = (2 + 2β2)π1 π(2+2β2)π₯ + (2 β 2β2)π2 π(2β2β2)π₯ 1 = (2 + 2β2)π1 + (2 β 2β2)π2 1 = (2 + 2β2)π1 + (2 β 2β2)(βπ1) 1 = π1(2 + 2β2 β 2 + 2β2) 1 = π1(4β2) π1 = 1 8 β2 π1 = βπ2 .................... (1) π2 = β 1 8 β2 Maka, π¦ = 1 8 β2π(2+2β2)π₯ β 1 8 β2π(2β2β2)π₯
5. 5. TASK II (February, 20th 2014) Solve these equation: 1. π4 π¦ ππ₯4 + 10 π2 π¦ ππ₯2 + 9π¦ = 0 2. π4 π¦ ππ₯4 + π3 π¦ ππ₯3 + π2 π¦ ππ₯2 + 2π¦ = 0 3. π¦β²β²β² + 4π¦β² = 0 4. π¦(4) + 4π¦β²β² β π¦β² + 6π¦ = 0 5. π6 π¦ ππ₯6 β 4 π5 π¦ ππ₯5 + 16 π4 π¦ ππ₯4 β 12 π3 π¦ ππ₯3 + 41 π2 π¦ ππ₯2 β 8 ππ¦ ππ₯ + 26π¦ = 0 Solution: 1. π4 π¦ ππ₯4 + 10 π2 π¦ ππ₯2 + 9π¦ = 0 Characteristic equation: π4 + 10π2 + 9 = 0 (π2 + 9)(π2 + 1) = 0 (π + 3π)(π β 3π)(π + π)(π β π) = 0 π1 = β3π β π2 = β3π β π3 = βπ β π3 = π So, the solution is π¦ = πΆ1 cos 3π₯ + πΆ2 sin 3π₯ + πΆ3 cos π₯ + πΆ4 sin π₯ 2. π4 π¦ ππ₯4 + π3 π¦ ππ₯3 + π2 π¦ ππ₯2 + 2π¦ = 0 Characteristic equation: π4 + π3 + π2 + 2 = 0 (π2 β π + 1)(π2 + 2π + 2) = 0 ο· (π2 β π + 1) = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π π1,2 = β(β1) Β± β(β1)2 β 4(1)(1) 2(1) π1,2 = 1 Β± β1 β 4 2 π1,2 = 1 Β± β3π 2 π1 = 1 2 + 1 2 β3π and π2 = 1 2 β 1 2 β3π ο· (π2 + 2π + 2) = 0 π3,4 = βπ Β± βπ2 β 4ππ 2π
6. 6. π3,4 = β(2) Β± β(2)2 β 4(1)(2) 2(1) π3,4 = β2 Β± β4 β 8 2 π3,4 = β2 Β± 2π 2 π3 = β1 + π and π4 = β1 β π So, the solution is π¦ = π 1 2 π₯ (πΆ1 cos 1 2 β3π₯ + πΆ2 sin 1 2 β3π₯) + πβπ₯ (πΆ3 cos π₯ + πΆ4 sin π₯) 3. π¦β²β²β² + 4π¦β² = 0 Characteristic equation: π3 + 4π = 0 π(π2 + 4) = 0 π(π + 2π)(π β 2π) = 0 π1 = 0 β π2 = β2π β π3 = 2π So, the solution is π¦ = πΆ1 + πΆ2cos2π₯ + πΆ3 sin 2π₯ 4. π¦(4) + 4π¦β²β² β π¦β² + 6π¦ = 0 Characteristics equation is π4 + 4π2 β π + 6 = 0 (π2 β π + 2)(π2 + π + 3) = 0 ο· (π2 β π + 2) = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π π1,2 = β(β1) Β± β(β1)2 β 4(1)(2) 2(1) π1,2 = 1 Β± β1 β 8 2 π1,2 = 1 Β± β7π 2 π1 = 1 2 + 1 2 β7π and π2 = 1 2 β 1 2 β7π ο· (π2 + π + 3) = 0 π3,4 = βπ Β± βπ2 β 4ππ 2π π3,4 = β(1) Β± β(1)2 β 4(1)(3) 2(1)
7. 7. π3,4 = β1 Β± β1 β 12 2 π3,4 = β1 Β± β11π 2 π3 = β 1 2 + 1 2 β11π and π4 = β 1 2 β 1 2 β11π So, the equation is: π¦ = π 1 2 π₯ (πΆ1 cos 1 2 β7π₯ + πΆ2 sin 1 2 β7π₯) + πβ 1 2 π₯ (πΆ3 cos 1 2 β11 π₯ + πΆ4 sin 1 2 β11 π₯) 5. π6 π¦ ππ₯6 β 4 π5 π¦ ππ₯5 + 16 π4 π¦ ππ₯4 β 12 π3 π¦ ππ₯3 + 41 π2 π¦ ππ₯2 β 8 ππ¦ ππ₯ + 26π¦ = 0 Characteristic equation is π6 β 4π5 + 16π4 β 12π3 + 41π2 β 8π + 26 = 0 (π4 + 3π2 + 2)(π2 β 4π + 13) = 0 (π2 + 1)(π2 + 2)(π2 β 4π + 13) = 0 ο· (π2 + 1) = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π π1,2 = 0 Β± β0 β 4(1)(1) 2(1) π1,2 = Β±ββ4 2 π1,2 = Β±2π 2 π1 = π and π2 = βπ ο· (π2 + 2) = 0 π3,4 = βπ Β± βπ2 β 4ππ 2π π3,4 = 0 Β± β0 β 4(1)(2) 2(1)
8. 8. π3,4 = Β±ββ8 2 π3,4 = Β±2β2π 2 π3 = β2π and π4 = ββ2π ο· (π2 β 4π + 13) = 0 π5,6 = βπ Β± βπ2 β 4ππ 2π π5,6 = β(β4) Β± β(β4)2 β 4(1)(13) 2(1) π5,6 = 4 Β± β16 β 52 2 π5,6 = 4 Β± 6π 2 π5 = 2 + 3π and π6 = 2 β 3π So, the solution is π¦ = πΆ1 cos π₯ + πΆ1 sin π₯ + π2π₯(πΆ3 cos 3π₯ + πΆ4 sin 3π₯) + πΆ5 cos β2π₯ + πΆ5 sin β2π₯
9. 9. TASK III (March 7th, 2014) 1. π3 π ππ‘3 β 5 π2 π ππ‘2 + 25 ππ ππ‘ β 125π = β60π7π‘ y(0)=0, yβ(0)=1, yβ(0)=2 2. π¦(πΌπ) β 6π¦β²β²β² + 16π¦β²β² + 54π¦β² β 225π¦ = 100πβ2π₯ π¦(0) = π¦β²(0) = π¦β²β²(0) = π¦β²β²β²(0) = 1 Solution: 1. π3 π ππ‘3 β 5 π2 π ππ‘2 + 25 ππ ππ‘ β 125π = β60π7π‘ y(0)=0, yβ(0)=1, yβ(0)=2 Quadratic equation: π3 β 5π2 + 25π β 125 = β60π7π‘ Y= yl + yr π3 β 5π2 + 25π β 125 = 0 (π β 5)(π + 5π)(π β 5π) = 0 π¦π = π1 π5π‘ + π2 cos 5π‘ + π3 sin 5π‘ π¦π = π΄0 π7π‘ π¦π β² = 7π΄0 π7π‘ π¦π β²β² = 49π΄0 π7π‘ π¦π β²β²β² = 343π΄0 π7π‘ π¦β²β²β² β 5π¦β²β² + 25π¦ β 125π¦ = β60π7π‘ 343π΄0 π7π‘ β 245π΄0 π7π‘ + 175π΄0 π7π‘ β 125π΄0 π7π‘ = β60π7π‘ 148π΄0 π7π‘ = β60π7π‘ π΄0 = β 60 148 = β 15 37 π¦π = β 15 37 π7π‘
10. 10. Then, we got the equation is equal to Y= yl + yr π¦(π‘) = π1 π5π‘ + π2 cos 5π‘ + π3 sin 5π‘ β 15 37 π7π‘ Now, we are going to find the value of c1, c2, and c3. y(0)=0 0 = π1 + π2 β 15 37 π1 + π2 = 15 37 ........................................ (1) yβ(0)=1 π¦β² (π‘) = 5π1 π5π‘ β 5π2 sin 5π‘ + 5π3 cos 5π‘ β 105 37 π7π‘ 1 = 5π1 + 5π3 β 105 37 π1 + π3 = 142 185 .................................... (2) yβ(0)=2 π¦β²β² (π‘) = 25π1 π5π‘ β 25π2 cos 5π‘ β 25π3 sin 5π‘ β 735 37 π7π‘ 2 = 25π1 β 25π2 β 735 37 25π1 β 25π2 = 809 37 π1 β π2 = 809 925 ....................................... (3) Elimination of π2 from (1) and (3) π1 β π2 = 809 925 ....................................... (3) π1 + π2 = 15 37 ........................................ (1) 2π1 = 809 + 375 925 π1 = 592 925 From (3), we can get the value of c2 by substituting the value of c1. π1 β π2 = 809 925 ....................................... (3) +
11. 11. 592 925 β π2 = 809 925 π2 = β 217 925 From (2), we can get the value of c3 by substituting the value of c1. π1 + π3 = 142 185 .................................... (2) 592 925 + π3 = 142 185 π3 = 118 925 After getting the value of c1, c2, and c3, then the equation is: π¦(π‘) = 592 925 π5π‘ β 217 925 cos 5π‘ + 118 925 sin 5π‘ β 15 37 π7π‘ 2. π¦(πΌπ) β 6π¦β²β²β² + 16π¦β²β² + 54π¦β² β 225π¦ = 100πβ2π₯ π¦(0) = π¦β²(0) = π¦β²β²(0) = π¦β²β²β²(0) = 1 Quadratic equation: π4 β 6π3 + 16π2 + 54π β 225 = 100πβ2π₯ π¦ = π¦π + π¦π π4 β 6π3 + 16π2 + 54π β 225 = 0 By using Horner method and ABC formula, so that the roots are gotten: π1 = 3 π3 = 3 + 4π π2 = β3 π4 = 3 β 4π π¦π = π1 π3π‘ + π2 πβ3π‘ + π3π‘ (π3 cos 4π‘ + π4 sin 4π‘) π¦π = π΄0 πβ2π₯ π¦π β² = β2π΄0 πβ2π‘
12. 12. π¦π β²β² = 4π΄0 πβ2π‘ π¦π β²β²β² = β8π΄0 πβ2π‘ π¦π (πΌπ) = 16π΄0 πβ2π‘ π¦(πΌπ) β 6π¦β²β²β² + 16π¦β²β² + 54π¦β² β 225π¦ = 100πβ2π‘ 16π΄0 πβ2π‘ + 48π΄0 πβ2π‘ + 64π΄0 πβ2π‘ β 108π΄0 πβ2π‘ β 225π΄0 πβ2π‘ = 100πβ2π‘ β205π΄0 πβ2π‘ = 100πβ2π‘ π΄0 = β 100 205 = β 20 41 So, the value of yr is π¦π = β 20 41 πβ2π‘ π¦ = π1 π3π‘ + π2 πβ3π‘ + π3π‘ (π3 cos 4π‘ + π4 sin 4π‘) β 20 41 πβ2π‘ Now, we are going to find the value of c1, c2, and c3. y(0) = 1 1 = π1 + π2 + π3 β 20 41 π1 + π2 + π3 = 61 41 ....................................... (1) yβ(0)=1 π¦β² = 3π1 π3π‘ β 3π2 πβ3π‘ + 3π3π‘(π3 cos 4π‘ + π4 sin 4π‘) + π3π‘(β4π3 sin 4π‘ + 4π4 cos 4π‘) + 40 41 πβ2π‘ 1 = 3π1 β 3π2 + 3π3 + 4π4 + 40 41 3π1 β 3π2 + 3π3 + 4π4 = 1 41 ........................ (2) yββ(0)=1
13. 13. π¦β²β² = 9π1 π3π‘ + 9π2 πβ3π‘ + 9π3π‘(π3 cos 4π‘ + π4 sin 4π‘) + 3π3π‘(β4π3 sin 4π‘ + 4π4 cos 4π‘) + 3π3π‘(β4π3 sin 4π‘ + 4π4 cos 4π‘) β 16π3π‘(π3 cos 4π‘ + π4 sin 4π‘) + 80 41 πβ2π‘ 1 = 9π1 + 9π2 + 9π3 + 12π4 + 12π4 β 16π3 β 80 41 1 = 9(π1 + π2 + π3) + 24π4 β 16π3 β 80 41 121 41 = 9( 61 41 ) + 24π4 β 16π3 β 428 41 = 24π4 β 16π3 2π3 β 3π4 = 107 82 .......................................... (3) yβββ(0)=1 π¦β²β²β² = 27π1 π3π‘ β 27π2 πβ3π‘ + 75π3π‘(π3 cos 4π‘ + π4 sin 4π‘) β 100π3π‘(βπ3 sin 4π‘ + π4 cos 4π‘) + 160 41 πβ2π‘ 1 = 27π1 β 27π2 + 75π3 β 100π4 + 160 41 27π1 β 27π2 + 75π3 β 100π4 = β 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: 27π1 β 27π2 + 75π3 β 100π4 = β 119 41 ........ (4) 27π1 β 27π2 + 27π3 + 36π4 = 9 41 ................ (2) 43π3 β 136π4 = β 128 41 6π3 β 176π4 = β 16 41 ..................................... (5) From (5) and (3), we can get the value of c4. +
14. 14. 6π3 β 176π4 = β 16 41 ..................................... (5) 6π3 β 9π4 = 321 82 ... ....................................... (3) β8π4 = β32 β 321 82 π4 = β353 656 From (3), we can get the value of c3. 2π3 β 3( β353 656 ) = 107 82 .......................................... (3) 2π3 = 1915 656 π3 = 1915 1315 From (1), we get: π1 + π2 + 1915 1315 = 61 41 ....................................... (1) π1 + π2 = 37 1312 ................................................ (6) From (2), we get: 3π1 β 3π2 + 3( 1915 1315 ) + 4( β353 656 ) = 1 41 ........................ (2) 3π1 β 3π2 = 32 β 5745 β 2824 1312 3π1 β 3π2 = β 8537 1312 π1 β π2 = β 8537 3936 ................................................ (7) From (6) and (7), we can get the value of c1 and c2 by elimination method. π1 + π2 = 37 1312 ................................................ (6) +
15. 15. π1 β π2 = β 8537 3936 ............................................. (7) 2π1 = β8537 + 111 3936 2π1 = β8426 3936 π1 = β4213 3936 From (7), we can get the value of c2. β4213 3936 β π2 = β 8537 3936 ............................................. (7) π2 = 8537 β 4213 3936 π2 = 1081 656 After getting the value of c1, c2, c3, and c4, we get the characteristic value. π¦ = β 4213 3936 π3π‘ + 1081 656 πβ3π‘ + π3π‘ ( 1915 1315 cos 4π‘ β 353 656 sin 4π‘) β 20 41 πβ2π‘ +
16. 16. TASK IV (March 13th, 2014) 1. π2 π ππ‘2 + 1000 ππ ππ‘ + 25000π = 24 π(0) = πβ²(0) = 0 2. π2 π¦ ππ‘2 β 4 ππ¦ ππ‘ + π¦ = 2π‘3 + 3π‘2 β 1 π¦(0) = π¦β²(0) = 1 Solution: 1. πβ²β² + 1000πβ² + 25000π = 24 Quadratic equation: π‘2 + 1000π‘ + 25000 = 0 π‘1,2 = βπ Β± βπ2 β 4ππ 2π π‘1,2 = β1000 Β± β(1000)2 β 4(1)(25000) 2(1) π‘1,2 = β1000 Β± β900000 2 π‘1,2 = β1000 Β± 300β10 2 π‘1,2 = β500 Β± 150β10 π‘1 = β500 + 150β10 and π‘2 = β500 β 150β10 ππ = π1 π(β500+150β10)π‘ + π2 π(β500β150β10)π‘ π π = π΄0 πβ² π = 0 πβ²β² π = 0 πβ²β² + 1000πβ² + 25000π = 24 0 + 1000(0) + 25000(π΄0) = 24 π΄0 = 3 3125 , then π π = 3 3125
17. 17. π = ππ + π π π(π‘) = π1 π(β500+150β10)π‘ + π2 π(β500β150β10)π‘ + 3 3125 π(0) = 0 0 = π1 + π2 + 3 3125 π1 + π2 = β 3 3125 ...................................(1) πβ²(π‘) = (β500 + 150β10)π1 π(β500+150β10)π‘ + (β500 β 150β10)π2 π(β500β150β10)π‘ πβ²(0) = 0 0 = (β500 + 150β10)π1 + (β500 β 150β10)π2 ........................... (2) By using elimination method, we can find the value of c1 from (1) and (2). (β500 β 150β10)π1 + (β500 β 150β10)π2 = β 3 3125 (β500 β 150β10) (β500 + 150β10)π1 + (β500 β 150β10)π2 = 0 β300β10π1 = β 3 3125 (β500 β 150β10) π1 = β5β10 β 15 3125 By using (1), we can find the value of c2. π1 + π2 = β 3 3125 β5β10 β 15 3125 + π2 = β 3 3125 π2 = 5β10 + 12 3125 After finding the value of c1 and c2, we get the equation. π(π‘) = ( β5β10 β 15 3125 ) π(β500+150β10)π‘ + ( 5β10 + 12 3125 ) π(β500β150β10)π‘ + 3 3125 β
18. 18. 2. π¦β²β² β 4π¦β² + π¦ = 2π‘3 + 3π‘2 β 1 Quadratic equation: π‘2 β 4π‘ + 1 = 0 π‘1,2 = βπ Β± βπ2 β 4ππ 2π π‘1,2 = β(β4) Β± β(β4)2 β 4(1)(1) 2(1) π‘1,2 = 4 Β± β12 2 π‘1,2 = 4 Β± 2β3 2 π‘1,2 = 2 Β± β3 π‘1 = 2 + β3 and π‘2 = 2 β β3 π¦π = π1 π(2+β3)π‘ + π2 π(2ββ3)π‘ π¦π = π΄3 π‘3 + π΄2 π‘2 + π΄1 π‘ + π΄0 π¦β² π = 3π΄3 π‘2 + 2π΄2 π‘ + π΄1 π¦β²β² π = 6π΄3 π‘ + 2π΄2 π¦β²β² β 4π¦β² + π¦ = 2π‘3 + 3π‘2 β 1 (6π΄3 π‘ + 2π΄2) β 4(3π΄3 π‘2 + 2π΄2 π‘ + π΄1) + (π΄3 π‘3 + π΄2 π‘2 + π΄1 π‘ + π΄0) = 2π‘3 + 3π‘2 β 1 π΄3 π‘3 + (β12π΄3 + π΄2)π‘2 + (6π΄3 β 8π΄2 + π΄1)π‘ + π΄1 + π΄0 = 2π‘3 + 3π‘2 β 1 Equation similarity: π΄3 = 2 β12π΄3 + π΄2 = 3 β12(2) + π΄2 = 3 π΄2 = 27 6π΄3 β 8π΄2 + π΄1 = 0 6(2) β 8(27) + π΄1 = 0 π΄1 = 204
19. 19. π΄1 + π΄0 = β1 204 + π΄0 = β1 π΄0 = β205 π¦π = 2π‘3 + 27π‘2 + 204π‘ β 205 π¦ = π¦π + π¦π π¦(π‘) = π1 π(2+β3)π‘ + π2 π(2ββ3)π‘ + 2π‘3 + 27π‘2 + 204π‘ β 205 π¦(0) = 1 1 = π1 + π2 β 205 π1 + π2 = 206 .........................(1) π¦β²(π‘) = (2 + β3)π1 π(2+β3)π‘ + (2 β β3)π2 π(2ββ3)π‘ + 6π‘2 + 54π‘ + 204 π¦β²(0) = 1 1 = (2 + β3)π1 + (2 β β3)π2 + 204 (2 + β3)π1 + (2 β β3)π2 = β203 ........................... (2) By using elimination and substitution method, the value of c1 and c2 can be obtained from (1) and (2). β203 = (2 + β3)π1 + (2 β β3)π2........................... (2) 206 = (2 β β3)π1 + (2 β β3)π2 ...................................(1) 2β3π1 = β203 β 206(2 β β3) 2β3π1 = β715 + 206β3 π1 = β715β3 + 618 6 π1 + π2 = 206 ( β715β3 + 618 6 ) + π2 = 206 π2 = 715β3 + 618 6 π¦(π‘) = ( β715β3+618 6 ) π(2+β3)π‘ + ( 715β3+618 6 ) π(2ββ3)π‘ + 2π‘3 + 27π‘2 + 204π‘ β 205 β
20. 20. TASK V (March 20th, 2014) 1. π2 π₯ ππ‘2 + 4 ππ₯ ππ‘ + 8π₯ = (20π‘2 + 16π‘ β 78)π2π‘ y(0)=yβ(0)=0 2. π3 π ππ‘3 β 5 π2 π ππ‘2 + 25 ππ ππ‘ β 125π = (β500π‘2 + 465π‘ β 387)π2π‘ q(0)=qβ(0)= qββ(0)=0 Solution: 1. π2 π₯ ππ‘2 + 4 ππ₯ ππ‘ + 8π₯ = (20π‘2 + 16π‘ β 78)π2π‘ y(0)=yβ(0)=0 quadratic equation is π2 + 4π + 8 = 0 π1,2 = βπ Β± βπ2 β 4ππ 2π π1,2 = β4 Β± β42 β 4(1)(8) 2(1) π1,2 = β4 Β± ββ16 2 π1,2 = β2 Β± 2π π¦π = πβ2π‘ (π1 cos 2π‘ + π2 sin 2π‘) π¦π = (π΄2 π‘2 + π΄1 π‘ + π΄0)π2π‘ π¦β² π = (2π΄2 π‘ + π΄1)π2π‘ + (π΄2 π‘2 + π΄1 π‘ + π΄0)(2π2π‘) π¦β²β² π = 2π΄2 π2π‘ + (2π΄2 π‘ + π΄1)(2π2π‘) + (2π΄2 π‘ + π΄1)(2π2π‘) + (π΄2 π‘2 + π΄1 π‘ + π΄0)(4π2π‘) π¦β²β² + 4π¦β² + 8π¦ = (20π‘2 + 16π‘ β 78)π2π‘ 2π΄2 π2π‘ + (2π΄2 π‘ + π΄1)(2π2π‘) + (2π΄2 π‘ + π΄1)(2π2π‘) + (π΄2 π‘2 + π΄1 π‘ + π΄0)(4π2π‘) + 4{(2π΄2 π‘ + π΄1)π2π‘ + (π΄2 π‘2 + π΄1 π‘ + π΄0)(2π2π‘)} + 8{(π΄2 π‘2 + π΄1 π‘ + π΄0)π2π‘} = (20π‘2 + 16π‘ β 78)π2π‘ (2π΄2 + 8π΄1 + 20π΄0)π2π‘ + (16π΄2 + 20π΄1)π‘π2π‘ + (20π΄2)π‘2 π2π‘ = (20π‘2 + 16π‘ β 78)π2π‘ ο· 20π΄2 = 20 π΄2 = 1
21. 21. ο· 16π΄2 + 20π΄1 = 16 16(1) + 20π΄1 = 16 20π΄1 = 0 π΄1 = 0 ο· 2π΄2 + 8π΄1 + 20π΄0 = β78 2(1) + 8(0) + 20π΄0 = β78 20π΄0 = β78 β 2 20π΄0 = β80 π΄0 = β4 π¦π = (π‘2 β 4)π2π‘ y = yl + yr π¦(π‘) = πβ2π‘(π1 cos 2π‘ + π2 sin 2π‘) + (π‘2 β 4)π2π‘ π¦β²(π‘) = (β2πβ2π‘)(π1 cos 2π‘ + π2 sin 2π‘) + πβ2π‘(β2π1 sin 2π‘ + 2π2 cos 2π‘) + 2π‘π2π‘ + (π‘2 β 4)(2π2π‘) π¦(0) = 0 0 = π1 β 4 π1 = 4 π¦β²(0) = 0 0 = β2π1 + 2π2 β 8 8 = β2(4) + 2π2 π2 = 8 π¦(π‘) = πβ2π‘(4 cos 2π‘ + 8 sin 2π‘) + (π‘2 β 4)π2π‘ 2. π3 π ππ‘3 β 5 π2 π ππ‘2 + 25 ππ ππ‘ β 125π = (β500π‘2 + 465π‘ β 387)π2π‘ q(0)=qβ(0)= qββ(0)=0 Quadratic equation is: π3 β 5π2 + 25π β 125 = 0
22. 22. π1 = 5 π2 = 5π π3 = β5π π¦π = π1 π5π‘ + π2 cos 5π‘ + π3 sin 5π‘ π¦π = (π΄2 π‘2 + π΄1 π‘ + π΄0)π2π‘ π¦β² π = (2π΄2 π‘ + π΄1)π2π‘ + (π΄2 π‘2 + π΄1 π‘ + π΄0)(2π2π‘ ) = 2π΄2 π‘π2π‘ + π΄1 π2π‘ + 2π΄2 π‘2 π2π‘ + 2π΄1 π‘π2π‘ + 2π΄0 π2π‘ π¦β²β² π = 2π΄2 π2π‘ + 8π΄2 π‘π2π‘ + 2π΄1 π2π‘ + 4π΄2 π‘π2π‘ + 4π΄2 π‘2 π2π‘ + 2π΄1 π2π‘ + 4π΄1 π‘π2π‘ + 4π΄0 π2π‘ π¦β²β²β² π = 12π΄2 π2π‘ + 12π΄1 π2π‘ + 8π΄0 π2π‘ + 24π΄2 π‘π2π‘ + 8π΄1 π‘π2π‘ + 8π΄2 π‘2 π2π‘ π¦β²β²β² β 5π¦β²β² + 25π¦β² β 125π¦ = (β500π‘2 + 465π‘ β 387)π2π‘ 12π΄2 π2π‘ + 12π΄1 π2π‘ + 8π΄0 π2π‘ + 24π΄2 π‘π2π‘ + 8π΄1 π‘π2π‘ + 8π΄2 π‘2 π2π‘ β 5{2π΄2 π2π‘ + 8π΄2 π‘π2π‘ + 2π΄1 π2π‘ + 4π΄2 π‘π2π‘ + 4π΄2 π‘2 π2π‘ + 2π΄1 π2π‘ + 4π΄1 π‘π2π‘ + 4π΄0 π2π‘ } + 25{2π΄2 π‘π2π‘ + π΄1 π2π‘ + 2π΄2 π‘2 π2π‘ + 2π΄1 π‘π2π‘ + 2π΄0 π2π‘} β 125{(π΄2 π‘2 + π΄1 π‘ + π΄0)π2π‘} = (β500π‘2 + 465π‘ β 387)π2π‘ (2π΄2 + 17π΄1 β 87π΄0)π2π‘ + (34π΄2 β 87π΄1)π‘π2π‘ + (β87π΄2)π‘2 π2π‘ = β500π‘2 π2π‘ + 465π‘π2π‘ β 387π2π‘ ο· β87π΄2 = β500 π΄2 = 500 87 ο· 34π΄2 β 87π΄1 = 465 34 ( 500 87 ) β 87π΄1 = 465 π΄1 = β23455 7569 ο· 2π΄2 + 17π΄1 β 87π΄0 = β387 2 ( 500 87 ) + 17 ( β23455 7569 ) β 87π΄0 = β387 π΄0 = 2617468 658503
23. 23. π¦π = ( 500 87 π‘2 β 23455 7569 π‘ + 2617468 658503 ) π2π‘ π¦ = π¦π + π¦π π¦(π‘) = π1 π5π‘ + π2 cos 5π‘ + π3 sin 5π‘ + ( 500 87 π‘2 β 23455 7569 π‘ + 2617468 658503 ) π2π‘ π¦β² (π‘) = 5π1 π5π‘ β 5π2 sin 5π‘ + 5π3 cos 5π‘ + ( 1000 87 π‘ β 23455 7569 ) π2π‘ + ( 500 87 π‘2 β 23455 7569 π‘ + 2617468 658503 ) (2π2π‘) π¦β²β²(π‘) = 25π1 π5π‘ β 25π2 cos5π‘ β 25π3 sin5π‘ + 1000 87 π2π‘ + ( 1000 87 π‘ β 23455 7569 ) (2π2π‘) + ( 1000 87 π‘ β 23455 7569 ) (2π2π‘) + ( 500 87 π‘2 β 23455 7569 π‘ + 2617468 658503 ) (4π2π‘) π¦(0) = 0 0 = π1 + π2 + 2617468 658503 π1 + π2 = β 2617468 658503 ................................... (1) π¦β²(0) = 0 0 = 5π1 + 5π3 β 23455 7569 + 5234936 658503 5π1 + 5π3 = β 3194351 658503 π1 + π3 = β 3194351 3292515 .................................... (2) π¦β²β²(0) = 0 0 = 25π1 β 25π2 + 1000 87 β 46910 7569 β 46910 7569 + 10469872 658503 25π1 β 25π2 = β 9876532 658503 .................................... (3) 25π1 + 25π2 = β 65436700 658503 ................................... (1) 50π1 = β 75313232 658503 π1 = β 75313232 32925150 +
24. 24. From equation (1), we get the value of c2. π2 = β 2617468 658503 + 75313232 32925150 π2 = β 55560168 32925150 From equation (2), we get the value of c3. π3 = β 3194351 3292515 + 75313232 32925150 π3 = 43369722 32925150 π¦(π‘) = β 75313232 32925150 π5π‘ β 55560168 32925150 cos 5π‘ + 43369722 32925150 sin 5π‘ + ( 500 87 π‘2 β 23455 7569 π‘ + 2617468 658503 ) π2π‘