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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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8. Answer:DFindthesolutiontothecharacteristicequationofeachanswerchoice,thentakethelimitas𝑡 → ∞

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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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11. Thisisaseparabledifferentialequation.

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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

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13.

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14.

A. 𝑢!! + 4𝑢! + 5𝑢 = 0B. 𝑢!! + 4𝑢! + 3𝑢 = 0C. 𝑢!! − 𝑢 = cos 𝑡D. 𝑢!! + 9𝑢 = 0E. 𝑢!! − 𝑢′− 6𝑢 = 0F. 𝑢!! + 2𝑢! + 𝑢 = 0G. 𝑢!! + 16𝑢 = 2 sin 4𝑡

a. Fb. Bc. Gd. De. A