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Ex 9.5, 10 - Show homogeneous: (1 + ex/y) dx + e x/y (1 - x/y)

Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 4
Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 5
Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 6
Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations - Part 7

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.