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Ex 9.6, 8 - Find general solution: (1 + x2) dy + 2xy dx - Ex 9.6

Ex 9.6, 8 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.6, 8 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.6, 8 - Chapter 9 Class 12 Differential Equations - Part 4

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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 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.