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Misc 4 - Prove x2 - y2 = c(x2 + y2)2 is general solution of

Misc 4 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 4 - Chapter 9 Class 12 Differential Equations - Part 3 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 4 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 5 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 6 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 7 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 8 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 9 Misc 4 - Chapter 9 Class 12 Differential Equations - Part 10

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Misc 3 Prove that ๐‘ฅ^2โˆ’๐‘ฆ^2=๐‘(๐‘ฅ^2+๐‘ฆ^2 )^2 is the general solution of differential equation (๐‘ฅ^3โˆ’3๐‘ฅ๐‘ฆ^2 )๐‘‘๐‘ฅ=(๐‘ฆ^3โˆ’3๐‘ฅ^2 ๐‘ฆ)๐‘‘๐‘ฆ, where ๐‘ is a parameter . Given differential equation (๐‘ฅ^3โˆ’3๐‘ฅ๐‘ฆ^2 )๐‘‘๐‘ฅ=(๐‘ฆ^3โˆ’3๐‘ฅ^2 ๐‘ฆ)๐‘‘๐‘ฆ (๐‘ฅ^3 โˆ’ 3๐‘ฅ๐‘ฆ^2)/(๐‘ฆ^3 โˆ’ 3๐‘ฅ^2 ๐‘ฆ)=๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘ฅ^3 โˆ’ 3๐‘ฅ๐‘ฆ^2)/(๐‘ฆ^(3 )โˆ’ 3๐‘ฅ^2 ๐‘ฆ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘ฅ^3 (1 โˆ’ (3๐‘ฅ๐‘ฆ^2)/๐‘ฅ^3 ))/(๐‘ฆ^(3 ) (1 โˆ’(3๐‘ฅ^2 ๐‘ฆ)/๐‘ฆ^3 ) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘ฅ^3 (1 โˆ’ (3๐‘ฆ^2)/๐‘ฅ^2 ))/(๐‘ฆ^(3 ) (1 โˆ’(3๐‘ฅ^2)/๐‘ฆ^2 ) ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=(๐‘ฅ/๐‘ฆ)^3ร—((1 โˆ’ 3(๐‘ฆ/๐‘ฅ)^2 ))/((1 โˆ’ 3(๐‘ฅ/๐‘ฆ)^2 ) ) Putting y = vx. Differentiating w.r.t. x ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ + ๐‘ฃ Putting value of ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ and y = vx in (1) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =(1/๐‘ฃ)^3ร—((1 โˆ’ 3๐‘ฃ^2 ))/((1 โˆ’ 3(1/๐‘ฃ)^2 ) ) โ€ฆ(1) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =1/๐‘ฃ^3 ร—((1 โˆ’ 3๐‘ฃ^2 ))/(((๐‘ฃ^2 โˆ’ 3)/๐‘ฃ^2 ) ) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ+๐‘ฃ =1/๐‘ฃร—((1 โˆ’ 3๐‘ฃ^2 ))/((๐‘ฃ^2 โˆ’ 3) ) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ=1/๐‘ฃร—((1 โˆ’ 3๐‘ฃ^2 ))/((๐‘ฃ^2 โˆ’ 3) )โˆ’๐‘ฃ ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ=1/๐‘ฃร—((1 โˆ’ 3๐‘ฃ^2 ) โˆ’ ๐‘ฃ ร— ๐‘ฃ (๐‘ฃ^2 โˆ’ 3))/((๐‘ฃ^2 โˆ’ 3) ) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ=1/๐‘ฃร—(1 โˆ’ 3๐‘ฃ^2 โˆ’ ๐‘ฃ^4 + 3๐‘ฃ^2)/((๐‘ฃ^2 โˆ’ 3) ) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ=1/๐‘ฃร—(1 โˆ’ ๐‘ฃ^4)/((๐‘ฃ^2 โˆ’ 3) ) ๐‘ฅ ๐‘‘๐‘ฃ/๐‘‘๐‘ฅ=(1 โˆ’ ๐‘ฃ^4)/((๐‘ฃ^3 โˆ’ 3๐‘ฃ) ) (๐‘ฃ^3 โˆ’3๐‘ฃ)๐‘‘๐‘ฃ/((1 โˆ’๐‘ฃ^4 ) )=๐‘‘๐‘ฅ/๐‘ฅ Integrating Both Sides โˆซ1โ–’ใ€–(๐‘ฃ^3 โˆ’3๐‘ฃ )/(1 โˆ’ ๐‘ฃ^4 ) ๐‘‘๐‘ฃใ€—=โˆซ1โ–’๐‘‘๐‘ฅ/๐‘ฅ โˆซ1โ–’ใ€–(๐‘ฃ^3 โˆ’3๐‘ฃ )/(1 โˆ’ ๐‘ฃ^4 ) ๐‘‘๐‘ฃใ€—=logโกใ€–|๐‘ฅ|ใ€—+๐ถ Let I = โˆซ1โ–’(๐‘ฃ^3 โˆ’ 3๐‘ฃ)/(1 โˆ’ ๐‘ฃ^4 ) ๐‘‘๐‘ฃ Therefore, ๐ผ =logโกใ€–|๐‘ฅ|+๐‘ใ€— โ€ฆ(1) Solving ๐‘ฐ ๐ผ =โˆซ1โ–’ใ€–(๐‘ฃ^3 โˆ’3๐‘ฃ )/(1 โˆ’ ๐‘ฃ^4 ) ๐‘‘๐‘ฃใ€— =โˆซ1โ–’ใ€–(๐‘ฃ^3 )/(1 โˆ’ ๐‘ฃ^4 )โˆ’3โˆซ1โ–’ใ€–๐‘ฃ/(1 โˆ’ใ€– ๐‘ฃใ€—^4 ) ๐‘‘๐‘ฃใ€—ใ€— ๐ผ =โˆซ1โ–’ใ€–๐‘ฃ^3/(โˆ’๐‘ก) ๐‘‘๐‘ก/(4๐‘ฃ^3 ) โˆ’3โˆซ1โ–’ใ€–๐‘ฃ/(1 โˆ’ ๐‘^2 ) ๐‘‘๐‘/2๐‘ฃใ€—ใ€— Put ๐‘ฃ^4โˆ’1=๐‘ก Diff. w.r.t. ๐‘ฃ ๐‘‘/๐‘‘๐‘ฃ (๐‘ฃ^4โˆ’1)=๐‘‘๐‘ก/๐‘‘๐‘ฃ 4๐‘ฃ^3=๐‘‘๐‘ก/๐‘‘๐‘ฃ ๐‘‘๐‘ฃ=๐‘‘๐‘ก/(4๐‘ฃ^3 ) Put ๐‘=๐‘ฃ^2 Diff. w.r.t. ๐‘ฃ ๐‘‘๐‘/๐‘‘๐‘ฃ=2๐‘ฃ ๐‘‘๐‘/2๐‘ฃ=๐‘‘๐‘ฃ โ€ฆ(2) ๐ผ =โˆ’1/4 โˆซ1โ–’ใ€– ๐‘‘๐‘ก/๐‘กโˆ’3/2 โˆซ1โ–’ใ€– ๐‘‘๐‘/(1 โˆ’ ๐‘^2 )ใ€—ใ€— ๐ผ =โˆ’1/4 โˆซ1โ–’ใ€– ๐‘‘๐‘ก/๐‘ก+3/2 โˆซ1โ–’ใ€– ๐‘‘๐‘/((๐‘^2 โˆ’ 1^2 ) )ใ€—ใ€— ๐ผ = (โˆ’1)/( 4) logโก๐‘ก+3/2 ร— 1/(2(1)) ๐‘™๐‘œ๐‘”((๐‘ โˆ’ 1)/(๐‘ + 1)) Putting t = ๐‘ฃ^4 โˆ’ 1 and p = v2 I = (โˆ’1)/4 logโกใ€–(๐‘ฃ^4โˆ’1) ใ€—+3/4 logโกใ€–((๐‘ฃ^2 โˆ’ 1))/((๐‘ฃ^2 + 1))ใ€— I = 1/4 [โˆ’logโกใ€–(๐‘ฃ^4โˆ’1)+3 ๐‘™๐‘œ๐‘” ((๐‘ฃ^2 โˆ’ 1))/((๐‘ฃ^2 + 1))ใ€— ] I = 1/4 [โˆ’logโกใ€–(๐‘ฃ^4โˆ’1)+ ๐‘™๐‘œ๐‘” (๐‘ฃ^2 โˆ’ 1)^3/(๐‘ฃ^2 + 1)^3 ใ€— ] I = 1/4 [๐‘™๐‘œ๐‘” (๐‘ฃ^2 โˆ’ 1)^3/(๐‘ฃ^2 + 1)^3 ร— 1/((๐‘ฃ^4 โˆ’ 1))] (โˆซ1โ–’ใ€–๐‘‘๐‘ฅ/(๐‘ฅ^2 โˆ’ ๐‘Ž^2 )=1/2๐‘Ž ๐‘™๐‘œ๐‘”((๐‘ฅ โˆ’ ๐‘Ž)/(๐‘ฅ + ๐‘Ž)) ใ€—) I = 1/4 [logโกใ€–1/((๐‘ฃ^4โˆ’1) )+ ๐‘™๐‘œ๐‘” (๐‘ฃ^2 โˆ’ 1)^3/(๐‘ฃ^2 + 1)^3 ใ€— ] I = 1/4 ๐‘™๐‘œ๐‘” 1/((๐‘ฃ^4 โˆ’ 1))ร—(๐‘ฃ^2 โˆ’ 1)^3/(๐‘ฃ^2 + 1)^3 I = 1/4 ๐‘™๐‘œ๐‘” 1/((๐‘ฃ^2 โˆ’ 1)(๐‘ฃ^2 + 1))ร—(๐‘ฃ^2 โˆ’ 1)^3/(๐‘ฃ^2 + 1)^3 I = 1/4 ๐‘™๐‘œ๐‘” (๐‘ฃ^2 โˆ’ 1)^2/(๐‘ฃ^2 + 1)^4 I = 1/4 logโกใ€–(((๐‘ฃ^2 โˆ’ 1))/(๐‘ฃ^2 + 1)^2 )^2 ใ€— I = 1/4 ร— 2 logโกใ€–((๐‘ฃ^2 โˆ’ 1))/(๐‘ฃ^2 + 1)^2 ใ€— I = 1/2 logโกใ€–((๐‘ฃ^2 โˆ’ 1))/(๐‘ฃ^2 + 1)^2 ใ€— Putting back v = ๐‘ฆ/๐‘ฅ I = 1/2 logโกใ€–(((๐‘ฆ/๐‘ฅ)^2 โˆ’ 1))/((๐‘ฆ/๐‘ฅ)^2 + 1)^2 ใ€— I = 1/2 log (((๐‘ฆ^2 โˆ’ ๐‘ฅ2)/๐‘ฅ^2 )/((๐‘ฆ^2 + ๐‘ฅ2)/๐‘ฅ^2 )^2 ) I = 1/2 log [(๐‘ฅ2(๐‘ฆ^2 โˆ’ ๐‘ฅ^2))/(๐‘ฆ^2 + ๐‘ฅ^2 )^2 ] Substituting value of I in (2) I = log |x| + c 1/2 log โŒˆ(๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/((๐‘ฆ2 + ๐‘ฅ2))โŒ‰ = log |x| + c log โŒˆ(๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/((๐‘ฅ2 + ๐‘ฆ2))โŒ‰ = 2 log |x| + 2c log โŒˆ(๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/((๐‘ฅ2 + ๐‘ฆ2))โŒ‰ = log |x|2 + log c1 log โŒˆ(๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/((๐‘ฅ2 + ๐‘ฆ2))โŒ‰ = log |x|2 + log c1 log โŒˆ(๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/(๐‘ฅ2 + ๐‘ฆ2)^2 โŒ‰ = log c1|x|2 Cancelling log (๐‘ฅ2(๐‘ฆ2 โˆ’ ๐‘ฅ2))/(๐‘ฅ2 + ๐‘ฆ2)^2 = c1 x2 x2 (y2 โˆ’ x2) = c1 x2 (x2 + y2)2 Cancelling x2 from both sides y2 โˆ’ x2 = c1 (x2 + y2)2 x2 โˆ’ y2 = c2 (x2 + y2)2 Hence proved

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.