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APM2611 ASSIGNMENT SOLUTIONS 2021 SEM1
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This document contains APM2611 Assignment 1 Solutions(Latest tutorial letter). All working are shown clearly and explanations are provided when necessary
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APM2611ASSIGNMENT 1 2021
QUESTION 1
QUESTION 1.1
4𝑧 ′′ − 4𝑧 ′ − 3𝑧 = cos(2𝑥)
𝐶𝑜𝑚𝑝𝑙𝑖𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑧𝑐 :
4𝑚2 − 4𝑚 − 3 = 0 ∴ 𝑐ℎ𝑎𝑟𝑎𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
(2𝑚 + 1)(2𝑚 − 3) = 0
1 3
𝑚1 = − 𝑎𝑛𝑑 𝑚2 =
2 2
𝑧𝑐 = 𝐶1 𝑒 𝑚1 𝑥 + 𝐶2 𝑒 𝑚2 𝑥
𝑥⁄ 3𝑥⁄
𝑧𝑐 = 𝐶1 𝑒 − 2 + 𝐶2 𝑒 2
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑧𝑝 :
𝐿𝑒𝑡: 𝑧𝑝 = 𝐴 + 𝐵 cos(2𝑥)
𝑧𝑝 = 𝐴 sin(2𝑥) + 𝐵 cos(2𝑥)
′
(𝑧𝑝 ) = 2𝐴 cos(2𝑥) − 2𝐵 sin(2𝑥)
′′
(𝑧𝑝 ) = −4𝐴 sin(2𝑥) − 4𝐵 cos(2𝑥)
4𝑧 ′′ − 4𝑧 ′ − 3𝑧 = cos(2𝑥)
′′ ′
4(𝑧𝑝 ) − 4(𝑧𝑝 ) − 3𝑧𝑝 = cos(2𝑥)
4[−4𝐴 sin(2𝑥) − 4𝐵 cos(2𝑥)] − 4[2𝐴 cos(2𝑥) − 2𝐵 sin(2𝑥)] − 3[𝐴 sin(2𝑥) + 𝐵 cos(2𝑥)]
= cos(2𝑥)
−16𝐴 sin(2𝑥) − 16𝐵 cos(2𝑥) − 8𝐴 cos(2𝑥) + 8𝐵 sin(2𝑥) − 3𝐴 sin(2𝑥) − 3𝐵 cos(2𝑥) = cos(2𝑥)
(−16𝐴 + 8𝐵 − 3𝐴) sin(2𝑥) + (−16𝐵 − 8𝐴 − 3𝐵) cos(2𝑥) = cos(2𝑥)
(−19𝐴 + 8𝐵) sin(2𝑥) + (−19𝐵 − 8𝐴) cos(2𝑥) = cos(2𝑥)
,−19𝐴 + 8𝐵 = 0 1
−8𝐴 − 19𝐵 = 1 2
19𝐴
𝐹𝑟𝑜𝑚 1 𝐵=
8
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝐵 𝑖𝑛𝑡𝑜 2
−8𝐴 − 19𝐵 = 1 2
19𝐴
−8𝐴 − 19 ( )=1
8
361𝐴
−8𝐴 − =1
8
−64𝐴 − 361𝐴 = 8
−425𝐴 = 8
8
𝐴=−
425
19𝐴
𝐵=
8
19 8 19
𝐵= (− )=−
8 425 425
𝑧𝑝 = 𝐴 sin(2𝑥) + 𝐵 cos(2𝑥)
8 19
𝑧𝑝 = − sin(2𝑥) − cos(2𝑥)
425 425
𝑧 = 𝑧𝑐 + 𝑧𝑝
𝒙⁄ 𝟑𝒙⁄ 𝟖 𝟏𝟗
𝒛 = 𝑪𝟏 𝒆− 𝟐 + 𝑪𝟐 𝒆 𝟐 − 𝐬𝐢𝐧(𝟐𝒙) − 𝐜𝐨𝐬(𝟐𝒙)
𝟒𝟐𝟓 𝟒𝟐𝟓
QUESTION 1.2
𝑥 2 𝑦 ′′ − 3𝑥𝑦 ′ + 3𝑦 = 12𝑥 4
, 3 3
𝑦 ′′ − 𝑦 ′ + 2 𝑦 = 12𝑥 4
𝑥 𝑥
𝐺𝑖𝑣𝑒𝑛 ∶ 𝑦1 = 𝑥 𝑎𝑛𝑑 𝑦2 = 𝑥 3
𝑦𝑝 = 𝑢1 𝑦1 + 𝑢2 𝑦2
𝑊 = |𝑥 𝑥 3 | = 3𝑥 3 − 𝑥 3 = 2𝑥 3
1 3𝑥 2
3
| 𝑥 2 𝑥 2 | 0 − 12𝑥 5
(𝑢1 )′ = 12𝑥 33𝑥 = = −6𝑥 2
2𝑥 2𝑥 3
−6𝑥 3
𝑢1 = ∫ −6𝑥 2 𝑑𝑥 = + 𝐶1 = −2𝑥 3 + 𝐶1
3
𝑥 0
| | 3 3
(𝑢2 )′ = 1 12𝑥 2 = 12𝑥 − 0 = 12𝑥 = 6
2𝑥 3 2𝑥 3 2𝑥 3
𝑢2 = ∫ 6 𝑑𝑥 = 6𝑥 + 𝐶2
𝑦𝑝 = 𝑢1 𝑦1 + 𝑢2 𝑦2
𝑦𝑝 = (−2𝑥 3 + 𝐶1 )𝑥 + (6𝑥 + 𝐶2 )𝑥 3
𝑦𝑝 = −2𝑥 4 + 𝐶1 𝑥 + 6𝑥 4 + 𝐶2 𝑥
𝒚𝒑 = 𝑪𝟏 𝒙 + 𝑪𝟐 𝒙 + 𝟒𝒙𝟒
QUESTION 2
𝑅𝑖𝑛 = 𝑖𝑛𝑝𝑢𝑡 𝑟𝑎𝑡𝑒
𝑅𝑜𝑢𝑡 = 𝑜𝑢𝑡𝑝𝑢𝑡 𝑟𝑎𝑡𝑒
𝑁 = 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠𝑎𝑙𝑡 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡.
𝑑𝑁
= 𝑅𝑖𝑛 − 𝑅𝑜𝑢𝑡
𝑑𝑡
1𝑔 4𝐿
𝑅𝑖𝑛 = × = 4 𝑔⁄𝑚𝑖𝑛
𝐿 𝑚𝑖𝑛
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