[SQP] Solve the differential equation: 𝑦+𝑑/𝑑π‘₯(π‘₯𝑦)=π‘₯(sin π‘₯+π‘₯) - 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: 𝑦+𝑑/𝑑π‘₯(π‘₯𝑦)=π‘₯(sin π‘₯+π‘₯)Now, our equation is 𝑦+𝑑/𝑑π‘₯(π‘₯𝑦)=π‘₯(sin π‘₯+π‘₯) Using product formula 𝑦+(𝑑(π‘₯))/𝑑π‘₯ 𝑦+π‘₯ 𝑑𝑦/𝑑π‘₯=π‘₯(sin π‘₯+π‘₯) 𝑦+𝑦+π‘₯ 𝑑𝑦/𝑑π‘₯=π‘₯(sin π‘₯+π‘₯) π‘₯ 𝑑𝑦/𝑑π‘₯+2𝑦=π‘₯(sin π‘₯+π‘₯) Diving both sides by x π‘₯/π‘₯ \ Γ— 𝑑𝑦/𝑑π‘₯+2𝑦/π‘₯=π‘₯(sin π‘₯+π‘₯)/π‘₯ 𝑑𝑦/𝑑π‘₯+2𝑦/π‘₯=(sin π‘₯+π‘₯) Comparing with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 𝟐/𝒙 & Q = (π’”π’Šπ’ 𝒙+𝒙) Finding Integrating factor (IF) IF = e^∫1▒𝑝𝑑π‘₯ = 𝒆^∫1β–’γ€–πŸ/𝒙 𝒅𝒙〗 = e^(2∫1β–’γ€–1/π‘₯ 𝑑π‘₯γ€—) = 𝒆^(𝟐 π’π’π’ˆβ‘|𝒙| ) = e^log⁑〖π‘₯^2 γ€— = 𝒙^𝟐 Solution of differential equation is y Γ— IF = ∫1▒〖𝑄.𝐼𝐹 𝑑π‘₯γ€— Putting values y Γ— x2 = ∫1β–’γ€–(π’”π’Šπ’ 𝒙+𝒙) 𝒙^𝟐 𝒅𝒙 γ€— yx2 = ∫1▒〖𝑠𝑖𝑛⁑π‘₯ Γ— π‘₯^2 γ€— 𝑑π‘₯+∫1▒𝒙^πŸ‘ 𝒅𝒙 yx2 = ∫1▒〖𝑠𝑖𝑛⁑π‘₯ Γ— π‘₯^2 γ€— 𝑑π‘₯+π‘₯^4/4+𝐢 yx2 = ∫1▒〖𝒙^𝟐 π’”π’Šπ’β‘π’™ γ€— 𝒅𝒙+𝒙^πŸ’/πŸ’+π‘ͺ Evaluating ∫1▒〖𝒙^𝟐 π’”π’Šπ’β‘π’™ γ€— 𝒅𝒙 separately ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— 𝑑π‘₯ We know that ∫1▒〖𝑓(π‘₯) 𝑔⁑(π‘₯) γ€— 𝑑π‘₯=𝑓(π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯βˆ’βˆ«1β–’(𝑓^β€² (π‘₯) ∫1▒𝑔(π‘₯) 𝑑π‘₯) 𝑑π‘₯ Putting f(x) = x2 and g(x) = sin x ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— 𝑑π‘₯=𝒙^𝟐 ∫1▒𝐬𝐒𝐧⁑𝒙 π’…π’™βˆ’βˆ«1β–’(𝒅(𝒙^𝟐 )/𝒅𝒙 ∫1β–’γ€–π’”π’Šπ’β‘π’™ 𝒅𝒙〗) 𝒅𝒙 = βˆ’ π‘₯^2 cos⁑π‘₯ βˆ’ ∫1β–’γ€–2π‘₯ Γ— βˆ’cos⁑〖π‘₯ 𝑑π‘₯γ€— γ€— = βˆ’ π‘₯^2 cos⁑π‘₯+2 ∫1▒〖𝒙 𝒄𝒐𝒔⁑𝒙 γ€— 𝒅𝒙+𝐢 Applying by parts again in ∫1▒〖𝒙 𝒄𝒐𝒔⁑𝒙 γ€— = βˆ’ π‘₯^2 cos⁑π‘₯+2 [π’™βˆ«1▒𝒄𝒐𝒔⁑𝒙 π’…π’™βˆ’βˆ«1β–’(𝒅𝒙/𝒅𝒙 ∫1▒〖𝒄𝒐𝒔⁑𝒙 𝒅𝒙〗) 𝒅𝒙]+𝐢 = βˆ’ π‘₯^2 cos⁑π‘₯+2 [𝒙 π’”π’Šπ’ π’™βˆ’βˆ«1β–’γ€–πŸ Γ— π’”π’Šπ’ 𝒙〗 𝒅𝒙]+𝐢 = βˆ’ π‘₯^2 cos⁑π‘₯+2 [π‘₯ 𝑠𝑖𝑛 π‘₯βˆ’βˆ«1β–’γ€–π’”π’Šπ’ 𝒙〗 𝒅𝒙]+𝐢 = βˆ’ π‘₯^2 cos⁑π‘₯+2 [π‘₯ 𝑠𝑖𝑛 π‘₯βˆ’(βˆ’π‘π‘œπ‘  π‘₯) ]+𝐢 = βˆ’ 𝒙^𝟐 𝒄𝒐𝒔⁑𝒙+𝟐 [𝒙 π’”π’Šπ’ 𝒙+πœπ¨π¬β‘π’™ ]+π‘ͺ Putting value of ∫1β–’γ€–π‘₯^2 sin⁑π‘₯ γ€— 𝑑π‘₯ in (1) yx2 = ∫1β–’γ€–π‘₯^2 𝑠𝑖𝑛⁑π‘₯ γ€— 𝑑π‘₯+π‘₯^4/4+𝐢 yx2 = βˆ’ 𝒙^𝟐 𝒄𝒐𝒔⁑𝒙+𝟐 [𝒙 π’”π’Šπ’ 𝒙+πœπ¨π¬β‘π’™ ]+𝒙^πŸ’/πŸ’+π‘ͺ 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