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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Example 21 Find the general solution of the differential equation 𝑦 𝑑π‘₯βˆ’(π‘₯+2𝑦^2 )𝑑𝑦=0 Given equation 𝑦 𝑑π‘₯βˆ’(π‘₯+2𝑦^2 )𝑑𝑦=0 𝑦 𝑑π‘₯=(π‘₯+2𝑦^2 )𝑑𝑦 𝑑𝑦/𝑑π‘₯ = 𝑦/(π‘₯ + 2𝑦^2 ) This is not of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 ∴ We find 𝑑π‘₯/𝑑𝑦 𝑑π‘₯/𝑑𝑦 = (π‘₯ + 2𝑦^2)/𝑦 𝑑π‘₯/𝑑𝑦 = (π‘₯ )/𝑦 + (2𝑦^2)/𝑦 𝑑π‘₯/𝑑𝑦 βˆ’ (π‘₯ )/𝑦 = 2y Differential equation is of the form 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 where P1 = (βˆ’1)/𝑦 & Q1 = 2y Now, IF = 𝑒^∫1▒〖𝑝_1 𝑑𝑦〗 IF = e^∫1β–’γ€–(βˆ’1)/𝑦 𝑑𝑦" " γ€— IF = e^(βˆ’log⁑𝑦 ) IF = e^log⁑(1/𝑦) IF = 1/𝑦 IF = e^log⁑(1/𝑦) IF = 1/𝑦 Solution is x(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑𝑦+𝐢 γ€— x Γ— 1/𝑦=∫1β–’γ€–2𝑦×1/𝑦〗 𝑑𝑦+𝐢 π‘₯/𝑦 = ∫1β–’γ€–2𝑑𝑦+𝐢〗 π‘₯/𝑦 = 2𝑦+𝐢 x = y (2y + C) x = 2y2 + Cy (As a log x = log xπ‘Ž) (As 𝑒^π‘™π‘œπ‘”β‘π‘₯ = x)

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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.