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Ex 9.4, 12 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : š‘„^2 š‘‘š‘¦+(š‘„š‘¦+š‘¦^2 ) š‘‘š‘„=0;š‘¦=1 When š‘„=1The differential equation can be written š‘Žs š‘„^2 š‘‘š‘¦ = āˆ’(xy + y2) dx š‘‘š‘¦/š‘‘š‘„ = (āˆ’(š‘„š‘¦ + š‘¦^2 ))/š‘„^2 "Let F(x, y) = " š‘‘š‘¦/š‘‘š‘„ " =" (āˆ’(š‘„š‘¦ +š‘¦^2 ))/š‘„^2 Finding F(š€x, š€y) F(šœ†x, šœ†y) = (āˆ’(šœ†š‘„šœ†š‘¦ + šœ†^2 š‘¦^2 ))/怖šœ†^2 š‘„怗^2 = (āˆ’šœ†^2 (š‘„š‘¦ + š‘¦^2 ))/怖šœ†^2 š‘„怗^2 = (āˆ’(š‘„š‘¦ + š‘¦^2 ))/š‘„^2 = šœ†Ā° F(x, y) = F(x, y) = (āˆ’(š‘„š‘¦ +š‘¦^2 ))/š‘„^2 āˆ“ F(x, y) is a homogenous function of degree zero Putting y = vx Diff w.r.t. x š’…š’š/š’…š’™ = x š’…š’—/š’…š’™ + v Putting value of š‘‘š‘¦/š‘‘š‘„ and y = vx in (1) š‘‘š‘¦/š‘‘š‘„ = (āˆ’(š‘„š‘¦ + š‘¦^2 ))/š‘„^2 v + (š’™ š’…š’—)/š’…š’™ = (āˆ’(š’™(š’—š’™) + (š’—š’™)^šŸ ))/š’™^šŸ v + (š‘„ š‘‘š‘£)/š‘‘š‘„ = (āˆ’(š‘„^2 š‘£ + š‘„^2 š‘£^2))/š‘„^2 v + (š‘„ š‘‘š‘£)/š‘‘š‘„ = āˆ’š‘„^2 ((š‘£ + š‘£^2))/š‘„^2 v + (š‘„ š‘‘š‘£)/š‘‘š‘„ = āˆ’(v2 + v) (š‘„ š‘‘š‘£)/š‘‘š‘„ = āˆ’ v2 āˆ’ v āˆ’ v (š‘„ š‘‘š‘£)/š‘‘š‘„ = āˆ’(š‘£^2+2š‘£) š’…š’—/(š’—^šŸ + šŸš’—) = āˆ’ š’…š’™/š’™ Integrating both sides āˆ«1ā–’š‘‘š‘£/(š‘£^2 + 2š‘£) = āˆ’āˆ«1ā–’š‘‘š‘„/š‘„ āˆ«1ā–’š‘‘š‘£/(š‘£^2 + 2š‘£) = āˆ’ log x + log c āˆ«1ā–’š‘‘š‘£/(怖(š‘£ć€—^2 + 2š‘£ + 1) āˆ’ 1) = āˆ’ log x + log c āˆ«1ā–’š’…š’—/((š’— + šŸ)^šŸ āˆ’ šŸ^šŸ ) = āˆ’ log x + log c šŸ/šŸ log (š’— + šŸ āˆ’ šŸ)/(š’— + šŸ + šŸ) = āˆ’ log x + log c 1/2 log š‘£/(š‘£ + 2) = āˆ’ log x + log c log āˆšš‘£/āˆš(š‘£ + 2) = āˆ’ log x + log C log āˆšš’—/āˆš(š’— + šŸ) + log x = log C log (š‘„ āˆšš‘£)/āˆš(š‘£ + 2) = log C Using āˆ«1ā–’š‘‘š‘„/(š‘„^2 āˆ’ š‘Ž^2 ) = 1/2š‘Ž log |(š‘„ āˆ’ š‘Ž)/(š‘„ + š‘Ž)|+š¶ (š‘„āˆšš‘£)/āˆš(š‘£ + 2) = C Putting value of v i.e š‘¦/š‘„ (š’™āˆš(š’š/š’™))/āˆš(š’š/š’™ + šŸ) = C āˆš(š‘„^2 Ɨ š‘¦/š‘„)/āˆš(š‘¦/š‘„ + 2) = C āˆšš‘„š‘¦/āˆš((š‘¦ + 2š‘„)/š‘„) = C (š‘„āˆšš‘¦)/āˆš(š‘¦ + 2š‘„) = C š‘„āˆšš‘¦ = Cāˆš(š‘¦+2š‘„) Squaring both sides x2y = c2(y + 2x) Putting x = 1 & y = 1 in (2) 12(1) = C2(1 + 2) 1 = 3C2 C2 = šŸ/šŸ‘ Putting value in (2) x2 y = 1/3(y + 2x) 3x2y = y + 2x y + 2x = 3x2y

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo