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Example 12 Show that the differential equation 2𝑦𝑒^(𝑥/𝑦) 𝑑𝑥+(𝑦−2𝑥𝑒^(𝑥/𝑦) )𝑑𝑦=0 is homogeneous and find its particular solution , given that, 𝑥=0 when 𝑦=1 2𝑦𝑒^(𝑥/𝑦) 𝑑𝑥+(𝑦−2𝑥𝑒^(𝑥/𝑦) )𝑑𝑦 = 0 Step 1: Finding 𝑑𝑥/𝑑𝑦 2𝑦𝑒^(𝑥/𝑦) 𝑑𝑥+(𝑦−2𝑥𝑒^(𝑥/𝑦) )𝑑𝑦=0 2𝑦𝑒^(𝑥/𝑦) 𝑑𝑥=−(𝑦−2𝑥𝑒^(𝑥/𝑦) )𝑑𝑦 2𝑦𝑒^(𝑥/𝑦) 𝑑𝑥=(2𝑥𝑒^(𝑥/𝑦)−𝑦)𝑑𝑦 Since the equation is in the form 𝑥/𝑦 , we will take 𝑑𝑥/𝑑𝑦 Instead of 𝑑𝑦/𝑑𝑥 𝒅𝒙/𝒅𝒚=((𝟐𝒙𝒆^(𝒙/𝒚) − 𝒚))/(𝟐𝒚𝒆^(𝒙/𝒚) ) Step 2: Put F(𝑥 , 𝑦)=𝑑𝑥/𝑑𝑦 and find F(𝜆𝑥 ,𝜆𝑦) F(𝑥 , 𝑦)= (2𝑥𝑒^(𝑥/𝑦) − 𝑦)/(2𝑦𝑒^(𝑥/𝑦) ) Finding F(𝝀𝒙 ,𝝀𝒚) F(𝜆𝑥 ,𝜆𝑦)=(2 (𝜆𝑥) 〖 𝑒〗^(𝜆𝑥/𝜆𝑦 −𝜆𝑦))/(2𝜆𝑦 〖 𝑒〗^(𝜆𝑥/𝜆𝑦 ) )=𝜆(2𝑥𝑒^(𝑥/𝑦) − 𝑦)/(𝜆 . 2𝑦 𝑒^(𝑥/𝑦) ) =(2𝑥𝑒^(𝑥/𝑦) − 𝑦)/(2𝑦 𝑒^(𝑥/𝑦) ) = F (𝒙 , 𝒚) So, F(𝜆𝑥 ,𝜆𝑦)= F(𝑥 , 𝑦) = 𝜆° F(𝑥 , 𝑦) Thus , F(𝑥 ,𝑦) is a homogeneous function of degree zero Therefore given differential equation is homogeneous differential equation Step 3: Solving 𝑑𝑥/𝑑𝑦 by Putting 𝑥=𝑣𝑦 𝑑𝑥/𝑑𝑦=(2𝑥 𝑒^(𝑥/𝑦) − 𝑦)/(2𝑦 𝑒^(𝑥/𝑦) ) Put 𝒙=𝒗𝒚 Diff. w.r.t. 𝑦 𝑑𝑥/𝑑𝑦=𝑑/𝑑𝑦 (𝑣𝑦) 𝑑𝑥/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 𝑑𝑦/𝑑𝑦 𝒅𝒙/𝒅𝒚=𝒚 . 𝒅𝒗/𝒅𝒚+𝒗 Putting values of 𝑑𝑥/𝑑𝑦 and x in (1) 𝑑𝑥/𝑑𝑦=(2𝑥𝑒^(𝑥/𝑦) − 𝑦)/(2𝑦 𝑒^(𝑥" " /𝑦) ) 𝒗+𝒚 𝒅𝒗/𝒅𝒚=(𝟐𝒗 𝒆^𝒗 − 𝟏)/(𝟐〖 𝒆〗^𝒗 ) 𝑦 𝑑𝑣/𝑑𝑦=(2𝑣 𝑒^𝑣 − 1)/(2〖 𝑒〗^𝑣 )−𝑣 (𝑦 𝑑𝑣)/𝑑𝑦=(2𝑣𝑒^𝑣 − 1 − 2𝑣𝑒^𝑣)/(2〖 𝑒〗^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−1)/(2〖 𝑒〗^𝑣 ) 𝟐〖 𝒆〗^𝒗 𝒅𝒗=(−𝒅𝒚)/( 𝒚) Integrating Both Sides ∫1▒〖2〖 𝑒〗^𝑣 𝑑𝑣〗=∫1▒(−𝑑𝑦)/( 𝑦) 𝟐〖 𝒆〗^𝒗=−𝐥𝐨𝐠⁡|𝒚|+𝒄 Putting back 𝑣=𝑥/𝑦 2𝑒^(𝑥/𝑦)=−𝑙𝑜𝑔|𝑦|+𝑐 2𝑒^(𝑥/𝑦)+𝑙𝑜𝑔|𝑦|=𝑐 Given that at 𝒙=𝟎 , 𝒚=𝟏 Putting 𝑥=0 and 𝑦=1 in (2) 2𝑒^(0/1)−𝑙𝑜𝑔|1|=𝑐 2 ×1+0=𝑐 𝒄=𝟐 Put Value of 𝑐 in (2) i.e., 2𝑒^(𝑥/𝑦)+𝑙𝑜𝑔|𝑦|=𝐶 𝟐𝒆^(𝒙/𝒚)+𝒍𝒐𝒈|𝒚|=𝟐" " is the particular solution of given differential equation

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.