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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 9.5, 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) π’…π’š/𝒅𝒙 + πŸπ’™/((𝟏 + π’™πŸ)) y = 𝒄𝒐𝒕⁑𝒙/((𝟏 + π’™πŸ)) Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q where P = πŸπ’™/((𝟏 + 𝒙^𝟐)) & Q = 𝒄𝒐𝒕⁑𝒙/((𝟏 + 𝒙^𝟐)) Finding Integrating factor, I.F I.F. = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€— I.F. = 𝒆^∫1β–’γ€–πŸπ’™/((𝟏 + 𝒙^𝟐 ) ) 𝒅𝒙 γ€— Let t = 1 + x2 dt = 2x dx I.F. = e^∫1▒〖𝑑𝑑/𝑑 γ€— = elog |t| = t Putting back t = (1 + x2) = (1 + x2) Solution of the equation is y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹.𝑑π‘₯+𝐢〗 Putting values, y.(1 + x2) = ∫1▒𝒄𝒐𝒕⁑𝒙/((𝟏 + π’™πŸ)) Γ— (𝟏+π’™πŸ).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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.