Question: (𝑥+2)𝑠𝑖𝑛(𝑦)+𝑥𝑐𝑜𝑠(𝑦)𝑦′=0 𝑢=𝑥𝑒𝑥
Solution:
𝑀=(𝑥+2)𝑠𝑖𝑛(𝑦) 𝑁=𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑐𝑜𝑠(𝑦) 𝑁𝑥=𝑐𝑜𝑠(𝑦)
therefore 𝑀𝑦≠𝑁𝑥 , therefore the equation is not exact.
multiplies the given integrating factor
(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑦′=0
𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦) 𝑁=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)
𝑀𝑦=(𝑥+2)𝑥𝑒𝑥𝑐𝑜𝑠(𝑦) 𝑁𝑥=𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦) + 2𝑥𝑒𝑥𝑐𝑜𝑠(𝑦)
then 𝑀𝑦=𝑁𝑥, therefore it becomes exact.
𝜙(𝑥,𝑦) s.t 𝜙𝑥=𝑀 𝜙𝑦=𝑁
𝜙=∫𝑥2𝑒𝑥𝑐𝑜𝑠(𝑦)𝑑𝑦
𝜙=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)
𝜙𝑥=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦) + 2𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ′(𝑥)=𝑀=(𝑥+2)𝑥𝑒𝑥𝑠𝑖𝑛(𝑦)+0
ℎ′(𝑥)=0
ℎ(𝑥)=𝑐
𝜙(𝑥,𝑦)=𝑥2𝑒𝑥𝑠𝑖𝑛(𝑦)+ℎ(𝑥)=𝑐
