Find the general solution of the following differential equation: š‘„ š‘‘š‘¦ āˆ’ (š‘¦ + 2š‘„ 2 )š‘‘š‘„ = 0

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Find general solution of differential equation: xdy - (y + 2x^2)dx = 0

Question 35 - CBSE Class 12 Sample Paper for 2021 Boards - Part 2
Question 35 - CBSE Class 12 Sample Paper for 2021 Boards - Part 3

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Question 35 Find the general solution of the following differential equation: š‘„ š‘‘š‘¦ āˆ’ (š‘¦ + 2š‘„2 )š‘‘š‘„ = 0 Given š‘„ š‘‘š‘¦ = (š‘¦ + 2š‘„2 )š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=(š‘¦ + 2š‘„^2)/š‘„ š‘‘š‘¦/š‘‘š‘„=š‘¦/š‘„+2š‘„ š’…š’š/š’…š’™āˆ’š’š/š’™=šŸš’™ Comparing with š’…š’š/š’…š’™ + Py = Q ∓ P = (āˆ’1)/š‘„ and Q = 2x Find integrating factor IF IF = e^∫1ā–’š‘ƒš‘‘š‘„ IF = š‘’^∫1▒〖(āˆ’1)/š‘„ š‘‘š‘„ć€— IF = š‘’^(āˆ’logā”š‘„ ) IF = š‘’^log⁔〖(š‘„)^(āˆ’1) 怗 IF = š‘’^怖log 〗⁔〖1/š‘„ć€— IF = šŸ/š’™ Solution of the equation y Ɨ I.F = ∫1ā–’ć€–š‘ø Ɨ š‘°.š‘­.š’…š’™+š’„ 怗 Putting values, š‘¦ Ɨ1/š‘„ = ∫1▒〖2š‘„ Ɨ1/š‘„ š‘‘š‘„ć€—+š¶ š‘¦/š‘„ = ∫1ā–’2š‘‘š‘„+š¶ š‘¦/š‘„ = 2š‘„+š¶ š’š = šŸš’™^šŸ+š‘Ŗš’™

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