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
