[SQP] Solve the differential equation: š‘¦+š‘‘/š‘‘š‘„(š‘„š‘¦)=š‘„(sin š‘„+š‘„) - CBSE Class 12 Sample Paper for 2026 Boards

part 2 - Question 34 (A) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12
part 3 - Question 34 (A) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 4 - Question 34 (A) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12 part 5 - Question 34 (A) - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards - Class 12

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Question 34 (A) Solve the differential equation: š‘¦+š‘‘/š‘‘š‘„(š‘„š‘¦)=š‘„(sin š‘„+š‘„)Now, our equation is š‘¦+š‘‘/š‘‘š‘„(š‘„š‘¦)=š‘„(sin š‘„+š‘„) Using product formula š‘¦+(š‘‘(š‘„))/š‘‘š‘„ š‘¦+š‘„ š‘‘š‘¦/š‘‘š‘„=š‘„(sin š‘„+š‘„) š‘¦+š‘¦+š‘„ š‘‘š‘¦/š‘‘š‘„=š‘„(sin š‘„+š‘„) š‘„ š‘‘š‘¦/š‘‘š‘„+2š‘¦=š‘„(sin š‘„+š‘„) Diving both sides by x š‘„/š‘„ \ Ɨ š‘‘š‘¦/š‘‘š‘„+2š‘¦/š‘„=š‘„(sin š‘„+š‘„)/š‘„ š‘‘š‘¦/š‘‘š‘„+2š‘¦/š‘„=(sin š‘„+š‘„) Comparing with š‘‘š‘¦/š‘‘š‘„ + Py = Q P = šŸ/š’™ & Q = (š’”š’Šš’ š’™+š’™) Finding Integrating factor (IF) IF = e^∫1ā–’š‘š‘‘š‘„ = š’†^∫1ā–’ć€–šŸ/š’™ š’…š’™ć€— = e^(2∫1▒〖1/š‘„ š‘‘š‘„ć€—) = š’†^(šŸ š’š’š’ˆā”|š’™| ) = e^logā”ć€–š‘„^2 怗 = š’™^šŸ Solution of differential equation is y Ɨ IF = ∫1ā–’ć€–š‘„.š¼š¹ š‘‘š‘„ć€— Putting values y Ɨ x2 = ∫1▒〖(š’”š’Šš’ š’™+š’™) š’™^šŸ š’…š’™ 怗 yx2 = ∫1ā–’ć€–š‘ š‘–š‘›ā”š‘„ Ɨ š‘„^2 怗 š‘‘š‘„+∫1ā–’š’™^šŸ‘ š’…š’™ yx2 = ∫1ā–’ć€–š‘ š‘–š‘›ā”š‘„ Ɨ š‘„^2 怗 š‘‘š‘„+š‘„^4/4+š¶ yx2 = ∫1ā–’ć€–š’™^šŸ š’”š’Šš’ā”š’™ 怗 š’…š’™+š’™^šŸ’/šŸ’+š‘Ŗ Evaluating ∫1ā–’ć€–š’™^šŸ š’”š’Šš’ā”š’™ 怗 š’…š’™ separately ∫1ā–’ć€–š‘„^2 sinā”š‘„ 怗 š‘‘š‘„ We know that ∫1ā–’ć€–š‘“(š‘„) š‘”ā”(š‘„) 怗 š‘‘š‘„=š‘“(š‘„) ∫1ā–’š‘”(š‘„) š‘‘š‘„āˆ’āˆ«1ā–’(š‘“^′ (š‘„) ∫1ā–’š‘”(š‘„) š‘‘š‘„) š‘‘š‘„ Putting f(x) = x2 and g(x) = sin x ∫1ā–’ć€–š‘„^2 sinā”š‘„ 怗 š‘‘š‘„=š’™^šŸ ∫1ā–’š¬š¢š§ā”š’™ š’…š’™āˆ’āˆ«1ā–’(š’…(š’™^šŸ )/š’…š’™ ∫1ā–’ć€–š’”š’Šš’ā”š’™ š’…š’™ć€—) š’…š’™ = āˆ’ š‘„^2 cosā”š‘„ āˆ’ ∫1▒〖2š‘„ Ɨ āˆ’cosā”ć€–š‘„ š‘‘š‘„ć€— 怗 = āˆ’ š‘„^2 cosā”š‘„+2 ∫1ā–’ć€–š’™ š’„š’š’”ā”š’™ 怗 š’…š’™+š¶ Applying by parts again in ∫1ā–’ć€–š’™ š’„š’š’”ā”š’™ 怗 = āˆ’ š‘„^2 cosā”š‘„+2 [š’™āˆ«1ā–’š’„š’š’”ā”š’™ š’…š’™āˆ’āˆ«1ā–’(š’…š’™/š’…š’™ ∫1ā–’ć€–š’„š’š’”ā”š’™ š’…š’™ć€—) š’…š’™]+š¶ = āˆ’ š‘„^2 cosā”š‘„+2 [š’™ š’”š’Šš’ š’™āˆ’āˆ«1ā–’ć€–šŸ Ɨ š’”š’Šš’ š’™ć€— š’…š’™]+š¶ = āˆ’ š‘„^2 cosā”š‘„+2 [š‘„ š‘ š‘–š‘› š‘„āˆ’āˆ«1ā–’ć€–š’”š’Šš’ š’™ć€— š’…š’™]+š¶ = āˆ’ š‘„^2 cosā”š‘„+2 [š‘„ š‘ š‘–š‘› š‘„āˆ’(āˆ’š‘š‘œš‘  š‘„) ]+š¶ = āˆ’ š’™^šŸ š’„š’š’”ā”š’™+šŸ [š’™ š’”š’Šš’ š’™+šœšØš¬ā”š’™ ]+š‘Ŗ Putting value of ∫1ā–’ć€–š‘„^2 sinā”š‘„ 怗 š‘‘š‘„ in (1) yx2 = ∫1ā–’ć€–š‘„^2 š‘ š‘–š‘›ā”š‘„ 怗 š‘‘š‘„+š‘„^4/4+š¶ yx2 = āˆ’ š’™^šŸ š’„š’š’”ā”š’™+šŸ [š’™ š’”š’Šš’ š’™+šœšØš¬ā”š’™ ]+š’™^šŸ’/šŸ’+š‘Ŗ Which is the required solution

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