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Last updated at Dec. 11, 2019 by Teachoo
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Example 17 Show that the differential equation 2๐ฆ๐^(๐ฅ/๐ฆ) ๐๐ฅ+(๐ฆโ2๐ฅ๐^(๐ฅ/๐ฆ) )๐๐ฆ=0 is homogeneous and find its particular solution , given that, ๐ฅ=0 when ๐ฆ=1 2๐ฆ๐^(๐ฅ/๐ฆ) ๐๐ฅ+(๐ฆโ2๐ฅ๐^(๐ฅ/๐ฆ) )๐๐ฆ = 0 Step 1: Find ๐๐ฅ/๐๐ฆ 2๐ฆ๐^(๐ฅ/๐ฆ) ๐๐ฅ+(๐ฆโ2๐ฅ๐^(๐ฅ/๐ฆ) )๐๐ฆ=0 2๐ฆ๐^(๐ฅ/๐ฆ) ๐๐ฅ=โ(๐ฆโ2๐ฅ๐^(๐ฅ/๐ฆ) )๐๐ฆ 2๐ฆ๐^(๐ฅ/๐ฆ) ๐๐ฅ=(2๐ฅ๐^(๐ฅ/๐ฆ)โ๐ฆ)๐๐ฆ Since the equation is in the form ๐ฅ/๐ฆ , we will take ๐๐ฅ/๐๐ฆ Instead of ๐๐ฆ/๐๐ฅ ๐๐ฅ/๐๐ฆ=((2๐ฅ๐^(๐ฅ/๐ฆ) โ ๐ฆ))/(2๐ฆ๐^(๐ฅ/๐ฆ) ) Step 2: Put F(๐ฅ , ๐ฆ)=๐๐ฅ/๐๐ฆ and find F(๐๐ฅ ,๐๐ฆ) F(๐ฅ , ๐ฆ)= (2๐ฅ๐^(๐ฅ/๐ฆ) โ ๐ฆ)/(2๐ฆ๐^(๐ฅ/๐ฆ) ) Finding F(๐๐ ,๐๐) F(๐๐ฅ ,๐๐ฆ)=(2 (๐๐ฅ) ใ ๐ใ^(๐๐ฅ/๐๐ฆ โ๐๐ฆ))/(2๐๐ฆ ใ ๐ใ^(๐๐ฅ/๐๐ฆ ) )=๐(2๐ฅ๐^(๐ฅ/๐ฆ) โ ๐ฆ)/(๐ . 2๐ฆ ๐^(๐ฅ/๐ฆ) ) =(2๐ฅ๐^(๐ฅ/๐ฆ) โ ๐ฆ)/(2๐ฆ ๐^(๐ฅ/๐ฆ) ) = F (๐ฅ , ๐ฆ) So, F(๐๐ฅ ,๐๐ฆ)= F(๐ฅ , ๐ฆ) = ๐ยฐ F(๐ฅ , ๐ฆ) Thus , F(๐ฅ ,๐ฆ) is a homogeneous function of degree zero Therefore given differential equation is homogeneous differential equation Step 3: Solving ๐๐ฅ/๐๐ฆ by Putting ๐ฅ=๐ฃ๐ฆ ๐๐ฅ/๐๐ฆ=(2๐ฅ ๐^(๐ฅ/๐ฆ) โ ๐ฆ)/(2๐ฆ ๐^(๐ฅ/๐ฆ) ) Put ๐ฅ=๐ฃ๐ฆ Diff. w.r.t. ๐ฆ ๐๐ฅ/๐๐ฆ=๐/๐๐ฆ (๐ฃ๐ฆ) ๐๐ฅ/๐๐ฆ=๐ฆ . ๐๐ฃ/๐๐ฆ+๐ฃ ๐๐ฆ/๐๐ฆ ๐๐ฅ/๐๐ฆ=๐ฆ . ๐๐ฃ/๐๐ฆ+๐ฃ Putting values of ๐๐ฅ/๐๐ฆ and x in (1) ๐๐ฅ/๐๐ฆ=(2๐ฅ๐^(๐ฅ/๐ฆ) โ ๐ฆ)/(2๐ฆ ๐^(๐ฅ" " /๐ฆ) ) ๐ฃ+๐ฆ ๐๐ฃ/๐๐ฆ=(2๐ฃ ๐^๐ฃ โ 1)/(2ใ ๐ใ^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(2๐ฃ ๐^๐ฃ โ 1)/(2ใ ๐ใ^๐ฃ )โ๐ฃ (๐ฆ ๐๐ฃ)/๐๐ฆ=(2๐ฃ๐^๐ฃ โ 1 โ 2๐ฃ๐^๐ฃ)/(2ใ ๐ใ^๐ฃ ) ๐ฆ ๐๐ฃ/๐๐ฆ=(โ1)/(2ใ ๐ใ^๐ฃ ) 2ใ ๐ใ^๐ฃ ๐๐ฃ=(โ๐๐ฆ)/( ๐ฆ) Integrating Both Sides โซ1โใ2ใ ๐ใ^๐ฃ ๐๐ฃใ=โซ1โ(โ๐๐ฆ)/( ๐ฆ) 2ใ ๐ใ^๐ฃ=โlogโก|๐ฆ|+๐ Putting back ๐ฃ=๐ฅ/๐ฆ 2๐^(๐ฅ/๐ฆ)=โ๐๐๐|๐ฆ|+๐ 2๐^(๐ฅ/๐ฆ)+๐๐๐|๐ฆ|=๐ โฆ(2) Given that at ๐ฅ=0 , ๐ฆ=1 Putting ๐ฅ=0 and ๐ฆ=1 in (2) 2๐^(0/1)โ๐๐๐|1|=๐ 2 ร1+0=๐ ๐=2 Put Value of ๐ in (2) i.e., 2๐^(๐ฅ/๐ฆ)+๐๐๐|๐ฆ|=๐ถ ๐๐^(๐/๐)+๐๐๐|๐|=๐" " is the particular solution of given differential equation (As ๐๐๐|1| = 0)
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