Author Topic: QUIZ2 TUT 0502  (Read 516 times)

Xinyu Jing

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QUIZ2 TUT 0502
« on: October 07, 2019, 12:12:42 AM »
Question: (3𝑥+6𝑦)+(𝑥2𝑦+3𝑦𝑥)𝑑𝑦𝑑𝑥=0

Solution: We want to find an integrating factor 𝜇 as a function of 𝑥𝑦 such that
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥, Let 𝑧=𝑥𝑦. Thus, 𝜇(𝑥𝑦)=𝜇(𝑧(𝑥,𝑦)) Then

𝜇𝑥(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑥=𝑦𝑑𝜇𝑑𝑧
𝜇𝑦(𝑥𝑦)=𝑑𝜇𝑑𝑧∂𝑧∂𝑦=𝑥𝑑𝜇𝑑𝑧

Therefore,
(𝜇𝑀)𝑦=(𝜇𝑁)𝑥

𝜇𝑀𝑦+𝑥𝑀𝑑𝜇𝑑𝑧=𝜇𝑁𝑥+𝑦𝑁𝑑𝜇𝑑𝑧

𝜇(𝑀𝑦−𝑁𝑥)=𝑑𝜇𝑑𝑧(𝑦𝑁−𝑥𝑀)

d𝜇d𝑧=𝜇(𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁)


Therefore,

𝜇(𝑧)=exp(∫𝑅(𝑧)d𝑧)
\quad where 𝑅(𝑧)=𝑅(𝑥𝑦)=𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁



𝑀(𝑥,𝑦)=3𝑥+𝑦 \quad and \quad 𝑁(𝑥,𝑦)=𝑥2𝑦+3𝑦𝑥=0

Then

∂∂𝑦𝑀(𝑥,𝑦)=−6𝑦2 \quad and \quad ∂∂𝑥𝑁(𝑥,𝑦)=2𝑥𝑦−3𝑦𝑥2

𝑁𝑥−𝑀𝑦𝑥𝑀−𝑦𝑁=2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥(3𝑥+6𝑦)−𝑦(𝑥2𝑦+3𝑦𝑥)
                     =2𝑥𝑦−3𝑦𝑥2+6𝑦22𝑥2+6𝑥𝑦−3𝑦2𝑥
                     =2𝑥𝑦−3𝑦𝑥2+6𝑦2𝑥𝑦(2𝑥𝑦−3𝑦𝑥2+6𝑦2)=1𝑥𝑦

Let 𝑥𝑦=𝑧

𝜇(𝑥𝑦)=exp(∫1𝑧d𝑧)=𝑒log|𝑧|=𝑧=𝑥𝑦

 
(3𝑥2𝑦+6𝑥)+(𝑥3+3𝑦2)𝑑𝑦𝑑𝑥=0


∂∂𝑦(3𝑥2𝑦+6𝑥)=3𝑥2=∂∂𝑥(𝑥3+3𝑦2)


𝜓𝑥(𝑥,𝑦)=3𝑥2𝑦+6𝑥(1)

𝜓𝑦(𝑥,𝑦)=𝑥3+3𝑦2(2)


𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+ℎ(𝑦)


𝜓𝑦(𝑥,𝑦)=𝑥3+ℎ′(𝑦)

Therefore,
ℎ′(𝑦)=3𝑦2

ℎ(𝑦)=𝑦3


𝜓(𝑥,𝑦)=𝑥3𝑦+3𝑥2+𝑦3


𝑥3𝑦+3𝑥2+𝑦3=𝐶