MAT1512 Assignment 8 2021

MAT1512 ASSIGNMENT 8 2021



Question 1



1.1
𝑑𝑦 𝑥 + 1
=
𝑑𝑥 𝑦 − 1
𝑑𝑦 (𝑥 + 1)
=
𝑑𝑥 (𝑦 − 1)


(𝑥 + 1)
𝑑𝑦 = 𝑑𝑥
(𝑦 − 1)
(𝑦 − 1)𝑑𝑦 = (𝑥 + 1) 𝑑𝑥



𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑒𝑝𝑒𝑟𝑎𝑏𝑙𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑛𝑙𝑦 𝑦 𝑖𝑠 𝑡ℎ𝑒
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑜𝑛𝑙𝑦 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.



1.2
𝑑𝑦 𝑦𝑒 𝑥+𝑦
=
𝑑𝑥 𝑥 2 + 2
𝑑𝑦 𝑦𝑒 𝑥 𝑒 𝑦
=
𝑑𝑥 𝑥 2 + 2
𝑦𝑒 𝑥 𝑒 𝑦
𝑑𝑦 = 𝑑𝑥
(𝑥 2 + 2)
𝑒𝑥
𝑑𝑦 = 𝑦𝑒 𝑦 𝑑𝑥
(𝑥 2 + 2)
𝑑𝑦 𝑒𝑥
= 𝑑𝑥
𝑦𝑒 𝑦 (𝑥 2 + 2)



𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑒𝑝𝑒𝑟𝑎𝑏𝑙𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑛𝑙𝑦 𝑦 𝑖𝑠 𝑡ℎ𝑒
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑜𝑛𝑙𝑦 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.

, 1.3
𝑑𝑦
= 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2
𝑑𝑥
𝑭𝒊𝒓𝒔𝒕 𝒑𝒐𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒕𝒚 (𝑻𝒓𝒆𝒂𝒕 𝒕𝒉𝒆 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒂𝒔 𝒈𝒊𝒗𝒆𝒏)



𝑁𝑜𝑡𝑒 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 ℎ𝑎𝑠 𝑥 𝑛𝑜𝑟 𝑦 𝑖𝑛 𝑖𝑡.
𝑇ℎ𝑖𝑠 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑤𝑖𝑡ℎ 𝑟𝑒𝑠𝑝𝑒𝑐𝑡 𝑡𝑜 𝑏𝑜𝑡ℎ 𝑥 𝑎𝑛𝑑 𝑦.
𝑑𝑦
= 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2
𝑑𝑥
𝑑𝑦 = [𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2 ]𝑑𝑥



𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑒𝑝𝑒𝑟𝑎𝑏𝑙𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑛𝑙𝑦 𝑦 𝑖𝑠 𝑡ℎ𝑒
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑜𝑛𝑙𝑦 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.


𝑑𝑦
= 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2
𝑑𝑥
𝑺𝒆𝒄𝒐𝒏𝒅 𝒑𝒐𝒔𝒔𝒊𝒃𝒊𝒍𝒊𝒕𝒚 (𝑨𝒔𝒔𝒖𝒎𝒆 𝒕𝒉𝒂𝒕 𝑺 𝒂𝒏𝒅 𝒕 𝒂𝒓𝒆 𝒕𝒉𝒆 𝒎𝒂𝒊𝒏 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆𝒔)
𝑑𝑆
= 𝑡(ln(𝑆 2𝑡 )) + 8𝑡 2
𝑑𝑡
𝑑𝑆
= 𝑡(ln(𝑆 2 )𝑡 ) + 8𝑡 2
𝑑𝑡
𝑑𝑆
= 𝑡(𝑡 ln(𝑆 2 )) + 8𝑡 2
𝑑𝑡
𝑑𝑆
= 𝑡 2 (ln(𝑆 2 )) + 8𝑡 2
𝑑𝑡
𝑑𝑆
= 𝑡 2 [ln(𝑆 2 ) + 8]
𝑑𝑡
𝑑𝑆 = 𝑡 2 [ln(𝑆 2 ) + 8]𝑑𝑡
𝑑𝑆
= 𝑡 2 𝑑𝑡
[ln(𝑆 2 ) + 8]



𝑇ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑠𝑒𝑝𝑒𝑟𝑎𝑏𝑙𝑒 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑜𝑛 𝑡ℎ𝑒 𝐿𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑛𝑙𝑦 𝑆 𝑖𝑠 𝑡ℎ𝑒
𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑎𝑛𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑅𝐻𝑆 𝑜𝑓 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛, 𝑜𝑛𝑙𝑦 𝑡 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒.