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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 11 Show that the differential equation 𝑥−𝑐𝑜𝑠(𝑦/𝑥)=𝑦 𝑐𝑜𝑠(𝑦/𝑥)+𝑥 is homogeneous and solve it.Step 1: Find 𝑑𝑦/𝑑𝑥 𝑥 𝑐𝑜𝑠(𝑦/𝑥) 𝑑𝑦/𝑑𝑥=𝑦 cos⁡(𝑦/𝑥)+𝑥 𝒅𝒚/𝒅𝒙=(𝒚 𝐜𝐨𝐬⁡ (𝒚/𝒙) + 𝒙)/(𝒙 𝐜𝐨𝐬⁡(𝒚/𝒙) ) Step 2: Put F(𝑥 ,𝑦)=𝑑𝑦/𝑑𝑥 & find F(𝜆𝑥 ,𝜆𝑦) F(𝑥 ,𝑦)=(𝑦 cos⁡ (𝑦/𝑥) + 𝑥)/(𝑥 cos⁡(𝑦/𝑥) ) Finding F(𝝀𝒙 ,𝝀𝒚) F(𝜆𝑥 ,𝜆𝑦)=((𝜆𝑦)𝑐𝑜𝑠(𝜆𝑦/𝜆𝑥) + 𝜆𝑥)/((𝜆𝑦) . cos⁡(𝜆𝑦/𝜆𝑥) ) =(𝜆𝑦 𝑐𝑜𝑠(𝑦/𝑥) + 𝜆𝑥)/(𝜆𝑦 cos⁡(𝑦/𝑥) ) =𝜆(𝑦 𝑐𝑜𝑠(𝑦/𝑥) + 𝑥)/(𝜆 𝑥 cos⁡(𝑦/𝑥) ) =(𝑦 𝑐𝑜𝑠(𝑦/𝑥) + 𝑥)/( 𝑥 cos⁡(𝑦/𝑥) ) = 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, the given differential equation is homogeneous differential equation Step 3: Solving 𝑑𝑦/𝑑𝑥 by Putting 𝑦=𝑣𝑥 𝑑𝑦/𝑑𝑥=(𝑦 𝑐𝑜𝑠(𝑦/𝑥) + 𝑥)/(𝑥 cos⁡(𝑦/𝑥) ) Put 𝒚=𝒗𝒙 So, 𝑑𝑦/𝑑𝑥=𝑑(𝑣𝑥) =𝑑𝑣/𝑑𝑥 . 𝑥+𝑣 𝑑𝑥/𝑑𝑥 =𝑑𝑣/𝑑𝑥 𝑥+𝑣 Putting values of 𝑑𝑦/𝑑𝑥 and y = vx in (1) i.e. 𝑑𝑦/𝑑𝑥 = (𝑦 𝑐𝑜𝑠(𝑦/𝑥)+𝑥)/(𝑥 cos⁡(𝑦/𝑥) ) 𝒅𝒗/𝒅𝒙 𝒙+𝒗=((𝒗𝒙) 𝒄𝒐𝒔(𝒗𝒙/𝒙) + 𝒙)/(𝒙 𝐜𝐨𝐬⁡(𝒗𝒙/𝒙) ) 𝑑𝑣/𝑑𝑥 𝑥+𝑣=(𝑣𝑥 𝑐𝑜𝑠(𝑣) + 𝑥)/(𝑥 cos⁡𝑣 ) 𝑑𝑣/𝑑𝑥 𝑥+𝑣=𝑥(𝑣 cos⁡〖𝑣 +1〗 )/(𝑥 cos⁡𝑣 ) 𝑑𝑣/𝑑𝑥 𝑥+𝑣=(𝑣 cos⁡〖𝑣 +1〗)/cos⁡𝑣 𝑑𝑣/𝑑𝑥 𝑥=(𝑣 cos⁡〖𝑣 + 1〗)/cos⁡𝑣 −𝑣 𝑑𝑣/𝑑𝑥 𝑥=(𝑣 cos⁡〖𝑣 + 1〗 −𝑣 cos⁡𝑣)/cos⁡𝑣 𝑑𝑣/𝑑𝑥 𝑥= 1/cos⁡𝑣 𝒄𝒐𝒔⁡𝒗 𝒅𝒗=𝒅𝒙/𝒙 Integrating Both Sides ∫1▒cos⁡〖𝑣 𝑑𝑣=∫1▒𝑑𝑥/𝑥〗 sin⁡〖𝑣=log⁡|𝑥|+𝑐1〗 Putting 𝒗=𝒚/𝒙 & t 𝒄𝟏=𝐥𝐨𝐠⁡𝒄 𝑠𝑖𝑛 𝑦/𝑥=log⁡〖|𝑥|+log⁡|𝑐| 〗 𝒔𝒊𝒏 𝒚/𝒙=𝒍𝒐𝒈⁡|𝒄𝒙|

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.