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Transcript

Misc 10 Solve the differential equation [𝑒^(βˆ’2√π‘₯)/√π‘₯βˆ’π‘¦/√π‘₯] 𝑑π‘₯/𝑑𝑦=1(π‘₯β‰ 0)[𝑒^(βˆ’2√π‘₯)/√π‘₯βˆ’π‘¦/√π‘₯] 𝑑π‘₯/𝑑𝑦=1 𝑒^(βˆ’2√π‘₯)/√π‘₯βˆ’ 𝑦/√π‘₯ =𝑑𝑦/𝑑π‘₯ π’…π’š/𝒅𝒙 + π’š/βˆšπ’™ = 𝒆^(βˆ’πŸβˆšπ’™)/βˆšπ’™ Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q where P = 𝟏/βˆšπ’™ & Q = 𝒆^(βˆ’πŸβˆšπ’™)/βˆšπ’™ IF. = e∫1β–’"pdx" Finding ∫1▒〖𝑷 𝒅𝒙〗 ∫1▒〖𝑷 𝒅𝒙=∫1▒𝒅𝒙/βˆšπ’™ γ€— ∫1▒〖𝑃 𝑑π‘₯=∫1β–’γ€–π‘₯^((βˆ’1)/2) 𝑑π‘₯γ€— γ€— ∫1▒〖𝑃 𝑑π‘₯=∫1β–’γ€–(π‘₯ (βˆ’1)/2 + 1)/((βˆ’1)/2 + 1) 𝑑π‘₯γ€— γ€— ∫1▒〖𝑃 𝑑π‘₯=2π‘₯^(1/2) γ€—= 2√π‘₯ ∴ IF = 𝒆^(πŸβˆšπ’™) Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 y𝒆^(πŸβˆšπ’™) = ∫1β–’γ€–(𝒆^(βˆ’πŸβˆšπ’™)/βˆšπ’™Γ—π’†^(πŸβˆšπ’™) )𝒅𝒙+𝒄〗 y𝑒^(2√π‘₯) = ∫1▒〖𝑑π‘₯/√π‘₯+𝑐〗 y𝑒^(2√π‘₯) = ∫1β–’γ€–1/√π‘₯ 𝑑π‘₯+𝑐〗 y𝑒^(2√π‘₯) = ∫1β–’γ€–π‘₯^((βˆ’1)/2) 𝑑π‘₯+𝑐〗 y𝑒^(2√π‘₯) = ∫1β–’γ€–(π‘₯^(βˆ’1/2 + 1) )/(βˆ’1/2 + 1)+𝑐〗 "y" 𝑒^(2√π‘₯) " = "2π‘₯^(1/2) + C "y" 𝒆^(πŸβˆšπ’™) " = "2βˆšπ’™ + C

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.