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

Transcript

Ex 9.5, 10 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. (1+๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฅ+๐‘’^(๐‘ฅ/๐‘ฆ) (1โˆ’๐‘ฅ/๐‘ฆ)๐‘‘๐‘ฆ=0 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 Step 1 : Find ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ (1+๐‘’^(๐‘ฅ/๐‘ฆ) )๐‘‘๐‘ฅ+๐‘’^(๐‘ฅ/๐‘ฆ) (1โˆ’๐‘ฅ/๐‘ฆ)๐‘‘๐‘ฆ = 0 (1+๐‘’^(๐‘ฅ/๐‘ฆ) ) dx = โˆ’๐‘’^(๐‘ฅ/๐‘ฆ) (1โˆ’๐‘ฅ/๐‘ฆ)๐‘‘๐‘ฆ ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ = (โˆ’๐‘’^(๐‘ฅ/๐‘ฆ) (1 โˆ’ ๐‘ฅ/๐‘ฆ) )/(1 + ๐‘’^(๐‘ฅ/๐‘ฆ) ) โ€ฆ(1) Step 2: Put ๐‘‘๐‘ฅ/๐‘‘๐‘ฆ = F(x, y) and find F(๐œ†x, ๐œ†y) F(x, y) = (โˆ’๐‘’^(๐‘ฅ/๐‘ฆ) (1 โˆ’ ๐‘ฅ/๐‘ฆ) )/(1 + ๐‘’^(๐‘ฅ/๐‘ฆ) ) F(๐œ†x, ๐œ†y) = (โˆ’๐‘’^(๐œ†๐‘ฅ/๐œ†๐‘ฆ) (1 โˆ’ ๐œ†๐‘ฅ/๐œ†๐‘ฆ) )/(1 + ๐‘’^(๐œ†๐‘ฅ/๐œ†๐‘ฆ) ) = (โˆ’๐‘’^(๐‘ฅ/๐‘ฆ) (1 โˆ’ ๐‘ฅ/๐‘ฆ) )/(1 + ๐‘’^(๐‘ฅ/๐‘ฆ) ) = ๐น(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 Putting 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=(โˆ’๐‘’^(๐‘ฅ/๐‘ฆ) (1 โˆ’ ๐‘ฅ/๐‘ฆ) )/(1 + ๐‘’^(๐‘ฅ/๐‘ฆ) ) ๐‘ฃ+๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃ (1 โˆ’ ๐‘ฃ))/(1 + ๐‘’^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃ (1 โˆ’ ๐‘ฃ))/(1 + ๐‘’^๐‘ฃ )โˆ’๐‘ฃ ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃ+ ๐‘ฃ๐‘’^๐‘ฃ)/(1 + ๐‘’^๐‘ฃ )โˆ’๐‘ฃ ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃ+ ๐‘ฃ๐‘’^๐‘ฃ โˆ’ ๐‘ฃ(1 + ๐‘’^๐‘ฃ ))/(1 + ๐‘’^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃ+ ๐‘ฃ๐‘’^๐‘ฃ โˆ’ ๐‘ฃ โˆ’ ๐‘ฃ๐‘’^๐‘ฃ)/(1 + ๐‘’^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’๐‘’^๐‘ฃโˆ’ ๐‘ฃ)/(1 + ๐‘’^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’(๐‘’^๐‘ฃ+ ๐‘ฃ))/(1 + ๐‘’^๐‘ฃ ) ๐‘ฆ ๐‘‘๐‘ฃ/๐‘‘๐‘ฆ=(โˆ’(๐‘’^๐‘ฃ+ ๐‘ฃ))/(1 + ๐‘’^๐‘ฃ ) ใ€–1 + ๐‘’ใ€—^๐‘ฃ/(๐‘ฃ + ๐‘’^๐‘ฃ ) ๐‘‘๐‘ฃ = (โˆ’๐‘‘๐‘ฆ)/๐‘ฆ Integrating both sides โˆซ1โ–’ใ€–ใ€–1 + ๐‘’ใ€—^๐‘ฃ/(๐‘ฃ + ๐‘’^๐‘ฃ ) ๐‘‘๐‘ฃ" " ใ€— =โˆซ1โ–’(โˆ’๐‘‘๐‘ฆ)/๐‘ฆ โˆซ1โ–’ใ€–ใ€–1 + ๐‘’ใ€—^๐‘ฃ/(๐‘ฃ + ๐‘’^๐‘ฃ ) ๐‘‘๐‘ฃใ€—=โˆ’logโกใ€–|๐‘ฆ|ใ€—+logโก๐‘ Putting v + ev = t (1 + ev) dv = dt Thus, our equation becomes โˆซ1โ–’๐‘‘๐‘ก/๐‘ก=โˆ’logโกใ€–|๐‘ฆ|ใ€—+logโก๐‘ logโกใ€–|๐‘ก|ใ€—=โˆ’logโกใ€–|๐‘ฆ|ใ€—+logโก๐‘ Putting back value of t = v + ev logโกใ€–|๐‘ฃ+๐‘’^๐‘ฃ |ใ€—=โˆ’logโกใ€–|๐‘ฆ|ใ€—+logโก๐‘ logโกใ€–|๐‘ฃ+๐‘’^๐‘ฃ |ใ€—+logโกใ€–|๐‘ฆ|ใ€—=logโก๐‘ logโก(|๐‘ฃ+๐‘’^๐‘ฃ |ร—|๐‘ฆ|)=logโก๐‘ logโก((๐‘ฃ+๐‘’^๐‘ฃ )ร—๐‘ฆ)=logโก๐‘ logโก(๐‘ฃ๐‘ฆ+๐‘’^๐‘ฃ ๐‘ฆ)=logโก๐‘ Putting back value of v = ๐‘ฅ/๐‘ฆ logโก(๐‘ฅ/๐‘ฆร—๐‘ฆ+๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘ฆ)=logโก๐‘ logโก(๐‘ฅ+๐‘’^(๐‘ฅ/๐‘ฆ) ๐‘ฆ)=logโก๐‘ Canceling log ๐’™+๐’š๐’†^(๐’™/๐’š)=๐‘ช

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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.