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Misc 18 - General solution: ex dy + (y ex + 2x) dx = 0 - Miscellaneous

Misc 18 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 18 - Chapter 9 Class 12 Differential Equations - Part 3


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Misc 18 The general solution of the differential equation 𝑒^π‘₯ 𝑑𝑦+(𝑦 𝑒^π‘₯+2π‘₯)𝑑π‘₯=0 is (A) π‘₯ 𝑒^𝑦+π‘₯^2=𝐢 (B) π‘₯ 𝑒^𝑦+𝑦^2=𝐢 (C) 𝑦 𝑒^π‘₯+π‘₯^2=𝐢 (D) 𝑦 𝑒^𝑦+π‘₯^2=𝐢 Given equation 𝑒^π‘₯ 𝑑𝑦+(𝑦 𝑒^π‘₯+2π‘₯)𝑑π‘₯=0 𝑒^π‘₯ 𝑑𝑦=βˆ’(𝑦 𝑒^π‘₯+2π‘₯)𝑑π‘₯ 𝑑𝑦/𝑑π‘₯= (βˆ’(𝑦𝑒^π‘₯ + 2π‘₯))/𝑒^π‘₯ 𝑑𝑦/𝑑π‘₯ = (βˆ’π‘¦π‘’^π‘₯)/𝑒^π‘₯ βˆ’2π‘₯/𝑒^π‘₯ 𝑑𝑦/𝑑π‘₯ = βˆ’π‘¦βˆ’2π‘₯/𝑒^π‘₯ 𝑑𝑦/𝑑π‘₯ + y = (βˆ’2π‘₯)/𝑒^π‘₯ Differential equation is of the form 𝑑𝑦/𝑑π‘₯ + Py = Q where P = 1 & Q = (βˆ’2π‘₯)/𝑒^π‘₯ IF = 𝑒^∫1▒〖𝑃 𝑑π‘₯γ€— IF = 𝑒^∫1β–’γ€–1 𝑑π‘₯γ€— IF = 𝑒^π‘₯ Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 yex = ∫1β–’γ€–(βˆ’2π‘₯)/𝑒^π‘₯ 𝑒^π‘₯ 𝑑π‘₯+𝑐〗 yex = βˆ’βˆ«1β–’γ€–2π‘₯ 𝑑π‘₯+𝑐〗 yex = βˆ’2Γ—π‘₯^2/2+𝑐 yex = βˆ’π‘₯^2+𝑐 yex + π‘₯^2=𝑐 ∴ Option (C) is the correct answer

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.