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Find the particular solution of the following differential equation, given that y = 0 when π‘₯ = πœ‹/4 𝑑𝑦/𝑑π‘₯+π‘¦π‘π‘œπ‘‘π‘₯ 2/(1 + sin⁑π‘₯ )

This question is similar to Example 22 - Chapter 9 Class 12 - Differential EquationsΒ 

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Question 8 (Choice 2) Find the particular solution of the following differential equation, given that y = 0 when π‘₯ = πœ‹/4 𝑑𝑦/𝑑π‘₯+π‘¦π‘π‘œπ‘‘π‘₯= 2/(1 + sin⁑π‘₯ ) 𝑑𝑦/𝑑π‘₯+π‘¦π‘π‘œπ‘‘π‘₯= 2/(1 + sin⁑π‘₯ ) Differential equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = cot x & Q = 𝟐/(𝟏 + π’”π’Šπ’β‘π’™ ) Now, IF = 𝑒^∫1▒〖𝑃 𝑑π‘₯γ€— IF = 𝑒^∫1β–’γ€–cot⁑π‘₯ 𝑑π‘₯γ€— IF = 〖𝑒^π₯𝐨𝐠⁑𝐬𝐒𝐧⁑𝒙 γ€—^" " IF = sin x Solution is y (IF) =∫1β–’γ€–(𝑄×𝐼𝐹) 𝑑π‘₯+𝑐〗 y sin x = ∫1β–’γ€–πŸ/(𝟏 + π’”π’Šπ’β‘π’™ )Γ— π’”π’Šπ’ 𝒙 𝒅𝒙〗 + C y sin x = 2∫1β–’γ€–(𝑠𝑖𝑛 π‘₯)/(1 + 𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯γ€— + C y sin x = 2∫1β–’γ€–((1 + 𝑠𝑖𝑛 π‘₯ βˆ’ 1))/(1 + 𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯γ€— + C y sin x = 2∫1β–’γ€–((1 + 𝑠𝑖𝑛 π‘₯))/(1 + 𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯γ€—βˆ’2∫1β–’γ€–1/(1 + 𝑠𝑖𝑛⁑π‘₯ ) 𝑑π‘₯γ€— + C y sin x = 𝟐∫1β–’π’…π’™βˆ’πŸβˆ«1β–’γ€–πŸ/(𝟏 + π’”π’Šπ’β‘π’™ ) 𝒅𝒙〗 + C y sin x = 2π‘₯βˆ’2∫1β–’γ€–πŸ/(𝟏 + π’”π’Šπ’β‘π’™ ) 𝒅𝒙〗 + C y sin x = 2π‘₯βˆ’2∫1β–’γ€–1/(1 + 𝑠𝑖𝑛⁑π‘₯ ) Γ—(𝟏 βˆ’ 𝐬𝐒𝐧⁑𝒙)/(𝟏 βˆ’ π’”π’Šπ’β‘π’™ ) 𝑑π‘₯γ€— + C y sin x = 2π‘₯βˆ’2∫1β–’γ€–(1 βˆ’ sin⁑π‘₯)/(1 βˆ’ sin^2⁑π‘₯ ) 𝑑π‘₯γ€— + C y sin x = 2π‘₯βˆ’2∫1β–’γ€–(𝟏 βˆ’ π’”π’Šπ’β‘π’™)/γ€–πœπ¨π¬γ€—^πŸβ‘π’™ 𝒅𝒙〗 + C y sin x = 2π‘₯βˆ’2[∫1β–’γ€–1/γ€–π‘π‘œπ‘ γ€—^2⁑π‘₯ 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–sin⁑π‘₯/γ€–π‘π‘œπ‘ γ€—^2⁑π‘₯ 𝑑π‘₯γ€—] + C y sin x = 2π‘₯βˆ’2[∫1β–’γ€–sec^2⁑π‘₯ 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–sin⁑π‘₯/(π‘π‘œπ‘  π‘₯) Γ—1/cos⁑π‘₯ 𝑑π‘₯γ€—] + C y sin x = 2π‘₯βˆ’2[∫1▒〖〖𝒔𝒆𝒄〗^πŸβ‘π’™ π’…π’™γ€—βˆ’βˆ«1β–’γ€–π­πšπ§β‘π’™ π¬πžπœβ‘π’™ 𝒅𝒙〗] + C y sin x = πŸπ’™βˆ’πŸ π­πšπ§β‘π’™+𝟐 𝒔𝒆𝒄 𝒙 + C We need to find particular solution when y = 0 when π‘₯ = πœ‹/4 Putting y = 0 and 𝒙 = 𝝅/πŸ’ 0 Γ— sin 𝝅/πŸ’ = 2(𝝅/πŸ’)βˆ’2 tan⁑〖𝝅/πŸ’γ€—+2 𝑠𝑒𝑐 𝝅/πŸ’ + C 0 = πœ‹/2βˆ’2 Γ— 1+2√2 + C 2βˆ’2√2βˆ’πœ‹/2 = C C = 𝟐(πŸβˆ’βˆšπŸ)βˆ’π…/𝟐 Thus, our particular solution is y sin x = πŸπ’™βˆ’πŸ π­πšπ§β‘π’™+𝟐 𝒔𝒆𝒄 𝒙 + 𝟐(πŸβˆ’βˆšπŸ)βˆ’π…/𝟐

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.