Question 21 (OR 2nd question) Find the general solution of the differential equation: 𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦 − 𝑥 tan𝑦 − 𝑥𝑦)/(𝑦 tan𝑦 )
𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦 − 𝑥 tan𝑦 − 𝑥𝑦)/(𝑦 tan𝑦 )
𝑑𝑥/𝑑𝑦=(𝑦 tan𝑦)/(𝑦 tan𝑦 )−(𝑥 tan𝑦 )/(𝑦 tan𝑦 )−𝑥𝑦/(𝑦 tan𝑦 )
𝑑𝑥/𝑑𝑦=1−𝑥/𝑦−𝑥/tan𝑦
𝑑𝑥/𝑑𝑦+𝑥/𝑦+𝑥/tan𝑦 =1
𝑑𝑥/𝑑𝑦+𝑥(1/𝑦+1/tan𝑦 )=1
Differential equation is of the form
𝑑𝑥/𝑑𝑦 + P1 x = Q1
where P1 = 1/𝑦+1/tan𝑦 & Q1 = 1
Now,
IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗
IF = e^∫1▒〖(1/𝑦 + 1/tan𝑦 )𝑑𝑦" " 〗
IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖1/tan𝑦 𝑑𝑦〗)
IF = e^(∫1▒〖1/𝑦 𝑑𝑦〗 +∫1▒〖cot𝑦 𝑑𝑦〗)
IF = e^(log𝑦 + logsin𝑦 )
IF = e^〖log 〗〖(𝑦 sin𝑦)〗
IF = y sin y
Solution is
x(IF) = ∫1▒〖(𝑄×𝐼𝐹)𝑑𝑦+𝐶 〗
x (y sin y) =∫1▒〖1×𝑦 sin𝑦 〗 𝑑𝑦+𝐶
xy sin y =∫1▒〖𝑦 sin𝑦 〗 𝑑𝑦+𝐶
(As ∫1▒cot𝑥 𝑑𝑥=logsin𝑥 )
(As log a + log b = log ab)
We know that
∫1▒〖𝑓(𝑦) 𝑔(𝑦) 〗 𝑑𝑦=𝑓(𝑦) ∫1▒𝑔(𝑦) 𝑑𝑦−∫1▒(𝑓^′ 𝑦∫1▒𝑔(𝑦) 𝑑𝑦) 𝑑𝑦
Putting f(y) = y and g(y) = sin y
xy sin y =𝑦" " ∫1▒sin𝑦 𝑑𝑦−∫1▒(𝑑(𝑦)/𝑑𝑦 ∫1▒〖sin𝑦 𝑑𝑦〗) 𝑑𝑦
xy sin y =−𝑦 cos𝑦 − ∫1▒〖−cos𝑦 𝑑𝑦〗
xy sin y =−𝑦 cos𝑦+ ∫1▒〖cos𝑦 𝑑𝑦〗
xy sin y =−𝑦 cos𝑦+sin𝑦+𝐶
xy sin y =sin𝑦−𝑦 cos𝑦+𝐶
x = (𝒔𝒊𝒏𝒚 − 𝒚 𝒄𝒐𝒔𝒚 + 𝑪" " )/(𝒚 𝐬𝐢𝐧𝒚 )

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.