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Last updated at Dec. 11, 2019 by Teachoo
Misc 4 Prove that ๐ฅ^2โ๐ฆ^2=๐(๐ฅ^2+๐ฆ^2 )^2 is the general solution of differential equation (๐ฅ^3โ3๐ฅ๐ฆ^2 )๐๐ฅ=(๐ฆ^3โ3๐ฅ^2 ๐ฆ)๐๐ฆ, where ๐ is a parameter . Given differential equation (๐ฅ^3โ3๐ฅ๐ฆ^2 )๐๐ฅ=(๐ฆ^3โ3๐ฅ^2 ๐ฆ)๐๐ฆ (๐ฅ^3 โ 3๐ฅ๐ฆ^2)/(๐ฆ^3 โ 3๐ฅ^2 ๐ฆ)=๐๐ฆ/๐๐ฅ ๐๐ฆ/๐๐ฅ=(๐ฅ^3 โ 3๐ฅ๐ฆ^2)/(๐ฆ^(3 )โ 3๐ฅ^2 ๐ฆ) ๐๐ฆ/๐๐ฅ=(๐ฅ^3 (1 โ (3๐ฅ๐ฆ^2)/๐ฅ^3 ))/(๐ฆ^(3 ) (1 โ(3๐ฅ^2 ๐ฆ)/๐ฆ^3 ) ) ๐๐ฆ/๐๐ฅ=(๐ฅ^3 (1 โ (3๐ฆ^2)/๐ฅ^2 ))/(๐ฆ^(3 ) (1 โ(3๐ฅ^2)/๐ฆ^2 ) ) ๐๐ฆ/๐๐ฅ=(๐ฅ/๐ฆ)^3ร((1 โ 3(๐ฆ/๐ฅ)^2 ))/((1 โ 3(๐ฅ/๐ฆ)^2 ) ) Putting y = vx. Differentiating w.r.t. x ๐๐ฆ/๐๐ฅ = ๐ฅ ๐๐ฃ/๐๐ฅ + ๐ฃ Putting value of ๐๐ฆ/๐๐ฅ and y = vx in (1) ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ =(1/๐ฃ)^3ร((1 โ 3๐ฃ^2 ))/((1 โ 3(1/๐ฃ)^2 ) ) โฆ(1) ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ =1/๐ฃ^3 ร((1 โ 3๐ฃ^2 ))/(((๐ฃ^2 โ 3)/๐ฃ^2 ) ) ๐ฅ ๐๐ฃ/๐๐ฅ+๐ฃ =1/๐ฃร((1 โ 3๐ฃ^2 ))/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร((1 โ 3๐ฃ^2 ))/((๐ฃ^2 โ 3) )โ๐ฃ ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร((1 โ 3๐ฃ^2 ) โ ๐ฃ ร ๐ฃ (๐ฃ^2 โ 3))/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร(1 โ 3๐ฃ^2 โ ๐ฃ^4 + 3๐ฃ^2)/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=1/๐ฃร(1 โ ๐ฃ^4)/((๐ฃ^2 โ 3) ) ๐ฅ ๐๐ฃ/๐๐ฅ=(1 โ ๐ฃ^4)/((๐ฃ^3 โ 3๐ฃ) ) (๐ฃ^3 โ3๐ฃ)๐๐ฃ/((1 โ๐ฃ^4 ) )=๐๐ฅ/๐ฅ Integrating Both Sides โซ1โใ(๐ฃ^3 โ3๐ฃ )/(1 โ ๐ฃ^4 ) ๐๐ฃใ=โซ1โ๐๐ฅ/๐ฅ โซ1โใ(๐ฃ^3 โ3๐ฃ )/(1 โ ๐ฃ^4 ) ๐๐ฃใ=logโกใ|๐ฅ|ใ+๐ถ Let I = โซ1โ(๐ฃ^3 โ 3๐ฃ)/(1 โ ๐ฃ^4 ) ๐๐ฃ Therefore, ๐ผ =logโกใ|๐ฅ|+๐ใ โฆ(1) Solving ๐ฐ ๐ผ =โซ1โใ(๐ฃ^3 โ3๐ฃ )/(1 โ ๐ฃ^4 ) ๐๐ฃใ =โซ1โใ(๐ฃ^3 )/(1 โ ๐ฃ^4 )โ3โซ1โใ๐ฃ/(1 โใ ๐ฃใ^4 ) ๐๐ฃใใ ๐ผ =โซ1โใ๐ฃ^3/(โ๐ก) ๐๐ก/(4๐ฃ^3 ) โ3โซ1โใ๐ฃ/(1 โ ๐^2 ) ๐๐/2๐ฃใใ Put ๐ฃ^4โ1=๐ก Diff. w.r.t. ๐ฃ ๐/๐๐ฃ (๐ฃ^4โ1)=๐๐ก/๐๐ฃ 4๐ฃ^3=๐๐ก/๐๐ฃ ๐๐ฃ=๐๐ก/(4๐ฃ^3 ) Put ๐=๐ฃ^2 Diff. w.r.t. ๐ฃ ๐๐/๐๐ฃ=2๐ฃ ๐๐/2๐ฃ=๐๐ฃ โฆ(2) ๐ผ =โ1/4 โซ1โใ ๐๐ก/๐กโ3/2 โซ1โใ ๐๐/(1 โ ๐^2 )ใใ ๐ผ =โ1/4 โซ1โใ ๐๐ก/๐ก+3/2 โซ1โใ ๐๐/((๐^2 โ 1^2 ) )ใใ ๐ผ = (โ1)/( 4) logโก๐ก+3/2 ร 1/(2(1)) ๐๐๐((๐ โ 1)/(๐ + 1)) Putting t = ๐ฃ^4 โ 1 and p = v2 I = (โ1)/4 logโกใ(๐ฃ^4โ1) ใ+3/4 logโกใ((๐ฃ^2 โ 1))/((๐ฃ^2 + 1))ใ I = 1/4 [โlogโกใ(๐ฃ^4โ1)+3 ๐๐๐ ((๐ฃ^2 โ 1))/((๐ฃ^2 + 1))ใ ] I = 1/4 [โlogโกใ(๐ฃ^4โ1)+ ๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ใ ] I = 1/4 [๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ร 1/((๐ฃ^4 โ 1))] (โซ1โใ๐๐ฅ/(๐ฅ^2 โ ๐^2 )=1/2๐ ๐๐๐((๐ฅ โ ๐)/(๐ฅ + ๐)) ใ) I = 1/4 [logโกใ1/((๐ฃ^4โ1) )+ ๐๐๐ (๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 ใ ] I = 1/4 ๐๐๐ 1/((๐ฃ^4 โ 1))ร(๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 I = 1/4 ๐๐๐ 1/((๐ฃ^2 โ 1)(๐ฃ^2 + 1))ร(๐ฃ^2 โ 1)^3/(๐ฃ^2 + 1)^3 I = 1/4 ๐๐๐ (๐ฃ^2 โ 1)^2/(๐ฃ^2 + 1)^4 I = 1/4 logโกใ(((๐ฃ^2 โ 1))/(๐ฃ^2 + 1)^2 )^2 ใ I = 1/4 ร 2 logโกใ((๐ฃ^2 โ 1))/(๐ฃ^2 + 1)^2 ใ I = 1/2 logโกใ((๐ฃ^2 โ 1))/(๐ฃ^2 + 1)^2 ใ Putting back v = ๐ฆ/๐ฅ I = 1/2 logโกใ(((๐ฆ/๐ฅ)^2 โ 1))/((๐ฆ/๐ฅ)^2 + 1)^2 ใ I = 1/2 log (((๐ฆ^2 โ ๐ฅ2)/๐ฅ^2 )/((๐ฆ^2 + ๐ฅ2)/๐ฅ^2 )^2 ) I = 1/2 log [(๐ฅ2(๐ฆ^2 โ ๐ฅ^2))/(๐ฆ^2 + ๐ฅ^2 )^2 ] Substituting value of I in (2) I = log |x| + c 1/2 log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฆ2 + ๐ฅ2))โ = log |x| + c log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = 2 log |x| + 2c log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = log |x|2 + log c1 log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/((๐ฅ2 + ๐ฆ2))โ = log |x|2 + log c1 log โ(๐ฅ2(๐ฆ2 โ ๐ฅ2))/(๐ฅ2 + ๐ฆ2)^2 โ = log c1|x|2 Cancelling log (๐ฅ2(๐ฆ2 โ ๐ฅ2))/(๐ฅ2 + ๐ฆ2)^2 = c1 x2 x2 (y2 โ x2) = c1 x2 (x2 + y2)2 Cancelling x2 from both sides y2 โ x2 = c1 (x2 + y2)2 x2 โ y2 = c2 (x2 + y2)2 Hence proved