APM2611 MAY JUNE 2019 MEMORANDUM
𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛 1
(1.1)
𝑦 ′′′ − 3𝑦 ′ + 2𝑦 = 0
𝐴𝑢𝑥𝑖𝑙𝑙𝑖𝑎𝑟𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑚3 − 3𝑚 + 2 = 0
𝐹𝑟𝑜𝑚 𝑡𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑒𝑟𝑟𝑜𝑟, 𝑚 = 1 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡.
(𝑚 − 1)(𝑚2 + 𝑚 − 2) = 0
(𝑚 − 1)(𝑚 + 2)(𝑚 − 1) = 0
𝑚 = −2 𝑜𝑟 𝑚 = 1 𝑜𝑟 𝑚 = 1
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑦𝑐 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝑦𝑝 = 𝐴
𝑦𝑝′ = 0
𝑦𝑝′′ = 0
𝑦𝑝′′′ = 0
𝑦 ′′′ − 3𝑦 ′ + 2𝑦 = 0
0 − 3(0) + 2𝐴 = 0
𝐴=0
,𝑦𝑝 = 0
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥 + 0
𝑦 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥
(1.2) 𝑈𝑛𝑑𝑒𝑟𝑡𝑒𝑚𝑖𝑛𝑒𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠
𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
𝐺𝑖𝑣𝑒𝑛 𝑦1 = 𝑐𝑜𝑠𝑥 𝑎𝑛𝑑 𝑦2 = 𝑠𝑖𝑛𝑥 𝑎𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
𝑆𝑜 𝑦𝐶 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝑦𝑝 = 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥
𝑦𝑝′ = −𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥
𝑦𝑝′′ = −𝐴𝑐𝑜𝑠𝑥 − 𝐵𝑠𝑖𝑛𝑥
𝑇𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑜𝑛 𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
−𝐴𝑐𝑜𝑠𝑥 − 𝐵𝑠𝑖𝑛𝑥 + 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥 = 𝑐𝑜𝑠𝑥
0 = 𝑐𝑜𝑠𝑥
𝑀𝑒𝑡ℎ𝑜𝑑 𝑖𝑠 𝑛𝑜𝑡 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑢𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑠 𝐴 𝑎𝑛𝑑 𝐵.
𝑇𝑜 𝑢𝑠𝑒 𝑦𝑝 = 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥
𝑦𝑝′ = 𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑖𝑛𝑥 + 𝐵𝑠𝑖𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥
, 𝑦𝑝′′ = −𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑐𝑜𝑠𝑥 + 𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥
𝑦𝑝′′ = −2𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 2𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥
𝑇𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑜𝑛 𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 2𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥 + 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 = 0𝑠𝑖𝑛𝑥 + 1𝑐𝑜𝑠𝑥
−2𝐴 = 0, 2𝐵 = 1
1
𝐴 = 0, 𝐵 =
2
𝑦𝑝 = 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥
1
𝑦𝑝 = 0𝑥𝑐𝑜𝑠𝑥 + 𝑥𝑠𝑖𝑛𝑥
2
1
𝑦𝑝 = 𝑥𝑠𝑖𝑛𝑥
2
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑦 = 𝑦𝑐 + 𝑦𝑝
1
𝑦 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥 + 𝑥𝑠𝑖𝑛𝑥
2
(1.3) 𝑉𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠
𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥, 𝑆𝑜 𝑓(𝑥) = 𝑐𝑜𝑠𝑥
𝐺𝑖𝑣𝑒𝑛 𝑦1 = 𝑐𝑜𝑠𝑥 𝑎𝑛𝑑 𝑦2 = 𝑠𝑖𝑛𝑥 𝑎𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
𝑆𝑜 𝑦𝐶 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥
𝑄𝑢𝑒𝑠𝑡𝑖𝑜𝑛 1
(1.1)
𝑦 ′′′ − 3𝑦 ′ + 2𝑦 = 0
𝐴𝑢𝑥𝑖𝑙𝑙𝑖𝑎𝑟𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
𝑚3 − 3𝑚 + 2 = 0
𝐹𝑟𝑜𝑚 𝑡𝑟𝑖𝑎𝑙 𝑎𝑛𝑑 𝑒𝑟𝑟𝑜𝑟, 𝑚 = 1 𝑖𝑠 𝑎 𝑟𝑜𝑜𝑡.
(𝑚 − 1)(𝑚2 + 𝑚 − 2) = 0
(𝑚 − 1)(𝑚 + 2)(𝑚 − 1) = 0
𝑚 = −2 𝑜𝑟 𝑚 = 1 𝑜𝑟 𝑚 = 1
𝐶𝑜𝑚𝑝𝑙𝑒𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝑦𝑐 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝑦𝑝 = 𝐴
𝑦𝑝′ = 0
𝑦𝑝′′ = 0
𝑦𝑝′′′ = 0
𝑦 ′′′ − 3𝑦 ′ + 2𝑦 = 0
0 − 3(0) + 2𝐴 = 0
𝐴=0
,𝑦𝑝 = 0
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑦 = 𝑦𝑐 + 𝑦𝑝
𝑦 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥 + 0
𝑦 = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 𝑥 + 𝑐3 𝑥𝑒 𝑥
(1.2) 𝑈𝑛𝑑𝑒𝑟𝑡𝑒𝑚𝑖𝑛𝑒𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠
𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
𝐺𝑖𝑣𝑒𝑛 𝑦1 = 𝑐𝑜𝑠𝑥 𝑎𝑛𝑑 𝑦2 = 𝑠𝑖𝑛𝑥 𝑎𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
𝑆𝑜 𝑦𝐶 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙
𝑦𝑝 = 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥
𝑦𝑝′ = −𝐴𝑠𝑖𝑛𝑥 + 𝐵𝑐𝑜𝑠𝑥
𝑦𝑝′′ = −𝐴𝑐𝑜𝑠𝑥 − 𝐵𝑠𝑖𝑛𝑥
𝑇𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑜𝑛 𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
−𝐴𝑐𝑜𝑠𝑥 − 𝐵𝑠𝑖𝑛𝑥 + 𝐴𝑐𝑜𝑠𝑥 + 𝐵𝑠𝑖𝑛𝑥 = 𝑐𝑜𝑠𝑥
0 = 𝑐𝑜𝑠𝑥
𝑀𝑒𝑡ℎ𝑜𝑑 𝑖𝑠 𝑛𝑜𝑡 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑢𝑠 𝑡𝑜 𝑓𝑖𝑛𝑑𝑖𝑛𝑔 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑠 𝐴 𝑎𝑛𝑑 𝐵.
𝑇𝑜 𝑢𝑠𝑒 𝑦𝑝 = 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥
𝑦𝑝′ = 𝐴𝑐𝑜𝑠𝑥 − 𝐴𝑥𝑠𝑖𝑛𝑥 + 𝐵𝑠𝑖𝑛𝑥 + 𝐵𝑥𝑐𝑜𝑠𝑥
, 𝑦𝑝′′ = −𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑐𝑜𝑠𝑥 + 𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥
𝑦𝑝′′ = −2𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 2𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥
𝑇𝑜 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑜𝑛 𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 − 𝐴𝑥𝑐𝑜𝑠𝑥 + 2𝐵𝑐𝑜𝑠𝑥 − 𝐵𝑥𝑠𝑖𝑛𝑥 + 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 = 𝑐𝑜𝑠𝑥
−2𝐴𝑠𝑖𝑛𝑥 + 2𝐵𝑐𝑜𝑠𝑥 = 0𝑠𝑖𝑛𝑥 + 1𝑐𝑜𝑠𝑥
−2𝐴 = 0, 2𝐵 = 1
1
𝐴 = 0, 𝐵 =
2
𝑦𝑝 = 𝐴𝑥𝑐𝑜𝑠𝑥 + 𝐵𝑥𝑠𝑖𝑛𝑥
1
𝑦𝑝 = 0𝑥𝑐𝑜𝑠𝑥 + 𝑥𝑠𝑖𝑛𝑥
2
1
𝑦𝑝 = 𝑥𝑠𝑖𝑛𝑥
2
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑦 = 𝑦𝑐 + 𝑦𝑝
1
𝑦 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥 + 𝑥𝑠𝑖𝑛𝑥
2
(1.3) 𝑉𝑎𝑟𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠
𝑦 ′′ + 𝑦 = 𝑐𝑜𝑠𝑥, 𝑆𝑜 𝑓(𝑥) = 𝑐𝑜𝑠𝑥
𝐺𝑖𝑣𝑒𝑛 𝑦1 = 𝑐𝑜𝑠𝑥 𝑎𝑛𝑑 𝑦2 = 𝑠𝑖𝑛𝑥 𝑎𝑟𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛𝑠.
𝑆𝑜 𝑦𝐶 = 𝑐1 𝑐𝑜𝑠𝑥 + 𝑐2 𝑠𝑖𝑛𝑥
