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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Example 17 Show that the differential equation 2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘‘๐‘ฅ+(๐‘ฆโˆ’2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฆ=0 is homogeneous and find its particular solution , given that, ๐‘ฅ=0 when ๐‘ฆ=1 2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘‘๐‘ฅ+(๐‘ฆโˆ’2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฆ = 0 Step 1: Find ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ 2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘‘๐‘ฅ+(๐‘ฆโˆ’2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฆ=0 2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘‘๐‘ฅ=โˆ’(๐‘ฆโˆ’2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฆ 2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘‘๐‘ฅ=(2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ)โˆ’๐‘ฆ)๐‘‘๐‘ฆ Since the equation is in the form ๐‘ฅ/๐‘ฆ , we will take ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ Instead of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=((2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ))/(2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ) Step 2: Put F(๐‘ฅ , ๐‘ฆ)=๐‘‘๐‘ฅ/๐‘‘๐‘ฆ and find F(๐œ†๐‘ฅ ,๐œ†๐‘ฆ) F(๐‘ฅ , ๐‘ฆ)= (2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ)/(2๐‘ฆ๐‘’^(๐‘ฅ/๐‘ฆ) ) Finding F(๐€๐’™ ,๐€๐’š) F(๐œ†๐‘ฅ ,๐œ†๐‘ฆ)=(2 (๐œ†๐‘ฅ) ใ€– ๐‘’ใ€—^(๐œ†๐‘ฅ/๐œ†๐‘ฆ โˆ’๐œ†๐‘ฆ))/(2๐œ†๐‘ฆ ใ€– ๐‘’ใ€—^(๐œ†๐‘ฅ/๐œ†๐‘ฆ ) )=๐œ†(2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ)/(๐œ† . 2๐‘ฆ ๐‘’^(๐‘ฅ/๐‘ฆ) ) =(2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ)/(2๐‘ฆ ๐‘’^(๐‘ฅ/๐‘ฆ) ) = F (๐‘ฅ , ๐‘ฆ) So, F(๐œ†๐‘ฅ ,๐œ†๐‘ฆ)= F(๐‘ฅ , ๐‘ฆ) = ๐œ†ยฐ F(๐‘ฅ , ๐‘ฆ) Thus , F(๐‘ฅ ,๐‘ฆ) is a homogeneous function of degree zero Therefore given differential equation is homogeneous differential equation Step 3: Solving ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ by Putting ๐‘ฅ=๐‘ฃ๐‘ฆ ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=(2๐‘ฅ ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ)/(2๐‘ฆ ๐‘’^(๐‘ฅ/๐‘ฆ) ) Put ๐‘ฅ=๐‘ฃ๐‘ฆ Diff. w.r.t. ๐‘ฆ ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=๐‘‘/๐‘‘๐‘ฆ (๐‘ฃ๐‘ฆ) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=๐‘ฆ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ+๐‘ฃ ๐‘‘๐‘ฆ/๐‘‘๐‘ฆ ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=๐‘ฆ . ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ+๐‘ฃ Putting values of ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ and x in (1) ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ=(2๐‘ฅ๐‘’^(๐‘ฅ/๐‘ฆ) โˆ’ ๐‘ฆ)/(2๐‘ฆ ๐‘’^(๐‘ฅ" " /๐‘ฆ) ) ๐‘ฃ+๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(2๐‘ฃ ๐‘’^๐‘ฃ โˆ’ 1)/(2ใ€– ๐‘’ใ€—^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(2๐‘ฃ ๐‘’^๐‘ฃ โˆ’ 1)/(2ใ€– ๐‘’ใ€—^๐‘ฃ )โˆ’๐‘ฃ (๐‘ฆ ๐‘‘๐‘ฃ)/๐‘‘๐‘ฆ=(2๐‘ฃ๐‘’^๐‘ฃ โˆ’ 1 โˆ’ 2๐‘ฃ๐‘’^๐‘ฃ)/(2ใ€– ๐‘’ใ€—^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’1)/(2ใ€– ๐‘’ใ€—^๐‘ฃ ) 2ใ€– ๐‘’ใ€—^๐‘ฃ ๐‘‘๐‘ฃ=(โˆ’๐‘‘๐‘ฆ)/( ๐‘ฆ) Integrating Both Sides โˆซ1โ–’ใ€–2ใ€– ๐‘’ใ€—^๐‘ฃ ๐‘‘๐‘ฃใ€—=โˆซ1โ–’(โˆ’๐‘‘๐‘ฆ)/( ๐‘ฆ) 2ใ€– ๐‘’ใ€—^๐‘ฃ=โˆ’logโก|๐‘ฆ|+๐‘ Putting back ๐‘ฃ=๐‘ฅ/๐‘ฆ 2๐‘’^(๐‘ฅ/๐‘ฆ)=โˆ’๐‘™๐‘œ๐‘”|๐‘ฆ|+๐‘ 2๐‘’^(๐‘ฅ/๐‘ฆ)+๐‘™๐‘œ๐‘”|๐‘ฆ|=๐‘ โ€ฆ(2) Given that at ๐‘ฅ=0 , ๐‘ฆ=1 Putting ๐‘ฅ=0 and ๐‘ฆ=1 in (2) 2๐‘’^(0/1)โˆ’๐‘™๐‘œ๐‘”|1|=๐‘ 2 ร—1+0=๐‘ ๐‘=2 Put Value of ๐‘ in (2) i.e., 2๐‘’^(๐‘ฅ/๐‘ฆ)+๐‘™๐‘œ๐‘”|๐‘ฆ|=๐ถ ๐Ÿ๐’†^(๐’™/๐’š)+๐’๐’๐’ˆ|๐’š|=๐Ÿ" " is the particular solution of given differential equation (As ๐‘™๐‘œ๐‘”|1| = 0)

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Davneet Singh
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.