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  1. Chapter 9 Class 12 Differential Equations
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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.