Check sibling questions

Example 21 - Find general solution: ydx - (x + 2y2)dy = 0

Example 21 - Chapter 9 Class 12 Differential Equations - Part 2
Example 21 - Chapter 9 Class 12 Differential Equations - Part 3


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 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.