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MAT3700 ASSIGNMENT 2 SEM 1 2023
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- University Of South Africa (Unisa)
This document contains MAT3700 ASSIGNMENT 2 SEM 1 2023 solutions. Clear step by step calculations are provided.
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MAT3700ASSIGNMENT 2
2023
, QUESTION 1
Solution:
𝑑2 𝑦 𝑑𝑦
+ 3 + 2𝑦 = 4𝑥 2
𝑑𝑥 2 𝑑𝑥
𝐴𝑢𝑥𝑖𝑙𝑖𝑎𝑟𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛: 𝑚2 + 3𝑚 + 2 = 0
𝑚2 + 3𝑚 + 2 = 0
(𝑚 + 1)(𝑚 + 2) = 0
𝑚1 = −1 𝑜𝑟 𝑚2 = −2
𝐶𝑜𝑚𝑝𝑙𝑖𝑚𝑒𝑛𝑡𝑎𝑟𝑦 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑦𝑐 = 𝐶1 𝑒 𝑚1 𝑥 + 𝐶2 𝑒 𝑚2 𝑥
𝑦𝑐 = 𝐶1 𝑒 𝑚1 𝑥 + 𝐶2 𝑒 𝑚2 𝑥
𝑦𝑐 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥
𝐵𝑦 𝑚𝑒𝑡ℎ𝑜𝑑 𝑜𝑓 𝑢𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒𝑑 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 ∶
𝑃𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑦𝑝 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶
𝑦𝑝 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶
𝑦 ′ 𝑝 = 2𝐴𝑥 + 𝐵
𝑦 ′′ 𝑝 = 2𝐴
𝑦 ′′ 𝑝 + 3𝑦 ′ 𝑝 + 2𝑦𝑝 = 4𝑥 2
2𝐴 + 3(2𝐴𝑥 + 𝐵) + 2(𝐴𝑥 2 + 𝐵𝑥 + 𝐶) = 4𝑥 2
, 2𝐴 + 6𝐴𝑥 + 3𝐵 + 2𝐴𝑥 2 + 2𝐵𝑥 + 2𝐶 = 4𝑥 2
2𝐴𝑥 2 + (6𝐴 + 2𝐵)𝑥 + (2𝐴 + 3𝐵 + 2𝐶) = 4𝑥 2
∴ 2𝐴 = 4 ⇒ 𝐴=2
∴ 6𝐴 + 2𝐵 = 0
6(2) + 2𝐵 = 0
12 + 2𝐵 = 0
2𝐵 = −12
𝐵 = −6
∴ 2𝐴 + 3𝐵 + 2𝐶 = 0
2(2) + 3(−6) + 2𝐶 = 0
4 − 18 + 2𝐶 = 0
−14 + 2𝐶 = 0
2𝐶 = 14
𝐶=7
𝑦𝑝 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶
𝑦𝑝 = 2𝑥 2 − 6𝑥 + 7
𝑦(𝑥) = 𝑦𝑐 + 𝑦𝑝
𝑦(𝑥) = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥 + 2𝑥 2 − 6𝑥 + 7
𝐺𝑒𝑛𝑒𝑟𝑎𝑙 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
𝑦(𝑥) = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −2𝑥 + 2𝑥 2 − 6𝑥 + 7
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