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

Transcript

Ex 9.6, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : (1+π‘₯^2 )𝑑𝑦+2π‘₯𝑦 𝑑π‘₯=cot⁑〖π‘₯ 𝑑π‘₯(π‘₯β‰ 0)γ€— Given equation (1 + x2)dy + 2xy dx = cot x dx Dividing both sides by dx (1 + x2)𝑑𝑦/𝑑π‘₯ + 2xy 𝑑π‘₯/𝑑π‘₯ = cot x 𝑑π‘₯/𝑑π‘₯ (1 + x2)𝑑𝑦/𝑑π‘₯ + 2xy = cot x Dividing both sides by (1 + x2) 𝑑𝑦/𝑑π‘₯ + 2π‘₯/((1 + π‘₯2)) y = cot⁑π‘₯/((1 + π‘₯2)) Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q where P = 2π‘₯/((1 + π‘₯^2)) & Q = cot⁑π‘₯/((1 + π‘₯^2)) Finding Integrating factor, I.F I.F. = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— I.F. = e^∫1β–’γ€–2π‘₯/((1 + π‘₯^2 ) ) 𝑑π‘₯ γ€— Let t = 1 + x2 dt = 2x dx …(1) I.F. = e^∫1▒〖𝑑𝑑/𝑑 γ€— = elog t = t = (1 + x2) So, I.F = 1 + x2 Solution of the equation is y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹.𝑑π‘₯+𝐢〗 Putting values, y.(1 + x2) = ∫1β–’cot⁑π‘₯/((1 + π‘₯2)) Γ— (1+π‘₯2).dx + c y.(1 + x2) = ∫1β–’γ€–cot⁑π‘₯ 𝑑π‘₯γ€—+𝐢 y (1 + x2) = log |sin⁑π‘₯ | + C Dividing by (1 + x2) y = (1 + x2)βˆ’1 log |𝐬𝐒𝐧⁑𝒙 |+π‘ͺ(𝟏+"x2" )^(βˆ’πŸ) is the general solution of the given equation Note: This answer does not match with the answer of the book. If we have made any mistake, please comment

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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.