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MATH251Exam1–SampleTestSolutions
1. a. Order:1
b. Linear:Yes
c. Separable:No;noamountofsimplifyingwillallowyoutoseparatethe𝑡’stoonesideof
theequationandthe𝑦’stotheothersideoftheequation.
d. Exact:Yes
𝑑𝑀𝑑𝑦 = 3𝑡!
𝑑𝑁𝑑𝑡 = 3𝑡!
2. Answer:C
3. Answer:B
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4. Answer:D
𝑆! = 700 𝑄! = 25 𝑟! = 3 𝑐! = 4 𝑟! = 2
5. Answer:BTofindthelimitingvelocity,set!"
!"= 0.
100−14 𝑣
! = 0
100 =14 𝑣
!𝑣! = 400𝑣 = 20
6. Answer:ARemembertogetyourequationinstandardformfirst.
7. Answer:B𝑦! = 2𝑒! ∙ 𝑒! = #𝑒! ,sobythePrincipleofSuperpositionanynumbertimes𝑒!isalsoasolutiontotheequation.
I. TrueII. Truebecause𝑦 = 0isasolutiontoeveryhomogenousequation.III. Truebecause𝑦!and𝑦!arenotconstantmultiplesofeachotherIV. Falsebecausethatisnotaconstantmultipleofanyoftheknownsolutions.
𝑆(𝑡) = 700 + (3 − 2)𝑡 = 700 + 𝑡
𝑄!(𝑡) = 12−2
700 + 𝑡 𝑄(𝑡)
𝑦!! + 3 tan 𝑡 𝑦! + 7 ln 𝑡 𝑦 = 0𝑝(𝑡) = 3 tan 𝑡 𝑊 = 𝐶𝑒!∫ ! !"# ! !"𝑊 = 𝐶𝑒!!!" (!"# !)𝑊 = 𝐶𝑒!" (!"#!! !)𝑊 = 𝐶𝑒!" (!"#! !)𝑊 = 𝐶cos! 𝑡
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9. Answer:B
𝑟! + 10𝑟 + 25 = 0
𝑟! + 5 +− 5 = 0
𝑟 = 0+ 5𝑖, 𝑟 = 0− 5𝑖,𝑟 = 0+ 5𝑖, 𝑟 = 0− 5𝑖
𝑦 = 𝑒!! cos 5𝑡 + 𝑒!! sin 5𝑡 + 𝑡𝑒!! cos 5𝑡 + 𝑡𝑒!! sin 5𝑡
𝑦 = cos 5𝑡 + sin 5𝑡 + 𝑡 cos 5𝑡 + 𝑡 sin 5𝑡
10. Answer:C
𝑀 = 2𝛼𝑦! − 2𝑦𝑒! + 4, 𝑁 = 2𝑥𝑦 − 𝑒!! − 3𝛼𝑦!
𝑑𝑀𝑑𝑦 =
𝑑𝑁𝑑𝑥
4𝛼𝑦 = 2𝑦
𝛼 =12
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12. a.
𝑀 = 4𝑥! + 2𝑦 + 𝑦 cos 𝑥 , 𝑁 = 2𝑥 + sin 𝑥 + 2𝑦
𝑑𝑀𝑑𝑦 = 2+ cos 𝑥
𝑑𝑁𝑑𝑥 = 2+ cos 𝑥
b. 𝜓 = 𝑀 𝑑𝑥 𝜓 = 𝑁 𝑑𝑦
𝜓 = 4𝑥! + 2𝑦 + 𝑦 cos 𝑥 𝑑𝑥 𝜓 = 2𝑥 + sin 𝑥 + 2𝑦 𝑑𝑦
𝜓 = 𝑥! + 2𝑦𝑥 + 𝑦 sin 𝑥 𝜓 = 2𝑥𝑦 + 𝑦 sin 𝑥 + 𝑦!
Sooverall𝜓 = 𝑥! + 2𝑥𝑦 + 𝑦 sin 𝑥 + 𝑦!
𝑥! + 2𝑥𝑦 + 𝑦 sin 𝑥 + 𝑦! = 𝐶
Use𝑦 0 = 3:0 ! + 2 0 3 + 3 sin 0 + 3 ! = 𝐶
𝐶 = 9
𝑥! + 2𝑥𝑦 + 𝑦 sin 𝑥 + 𝑦! = 9