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

Transcript

Misc 4 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 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.