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Ex 9.4, 4 - Find general solution: sec2 x tan y dx + sec2y

Ex 9.4, 4 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.4, 4 - Chapter 9 Class 12 Differential Equations - Part 3

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Ex 9.4, 4 For each of the differential equations in Exercises 1 to 10, find the general solution : sec^2⁑〖π‘₯ tan⁑〖𝑦 𝑑π‘₯+sec^2⁑〖𝑦 tan⁑〖π‘₯𝑑𝑦=0γ€— γ€— γ€— γ€— sec^2⁑〖π‘₯ tan⁑〖𝑦 𝑑π‘₯+sec^2⁑〖𝑦 tan⁑〖π‘₯𝑑𝑦=0γ€— γ€— γ€— γ€— Dividing both sides by tan y tan x (𝑠𝑒𝑐2π‘₯ π‘‘π‘Žπ‘›β‘π‘¦ 𝑑π‘₯ +〖𝑠𝑒𝑐〗^2⁑𝑦 π‘‘π‘Žπ‘›β‘γ€–π‘₯ 𝑑𝑦〗)/π‘‘π‘Žπ‘›β‘γ€–π‘¦ π‘‘π‘Žπ‘›β‘π‘₯ γ€— = 0/π‘‘π‘Žπ‘›β‘γ€–π‘₯ π‘‘π‘Žπ‘›β‘π‘¦ γ€— (𝑠𝑒𝑐2π‘₯ π‘‘π‘Žπ‘›π‘¦ 𝑑π‘₯)/π‘‘π‘Žπ‘›β‘γ€–π‘¦ π‘‘π‘Žπ‘›β‘π‘₯ γ€— + (𝑠𝑒𝑐2𝑦 π‘‘π‘Žπ‘›π‘₯ 𝑑𝑦)/π‘‘π‘Žπ‘›β‘γ€–π‘¦ π‘‘π‘Žπ‘›β‘π‘₯ γ€— = 0 (𝑠𝑒𝑐2π‘₯ )/tan⁑π‘₯ dx + 𝑠𝑒𝑐2𝑦/tan⁑𝑦 dy = 0 Integrating both sides ∫1β–’γ€–(𝑠𝑒𝑐2π‘₯/tan⁑π‘₯ 𝑑π‘₯+𝑠𝑒𝑐2𝑦/tan⁑𝑦 𝑑𝑦)=γ€— 0 ∫1▒〖𝑠𝑒𝑐2π‘₯/tan⁑π‘₯ 𝑑π‘₯+∫1▒〖𝑠𝑒𝑐2𝑦/tan⁑𝑦 𝑑𝑦〗=γ€— 0 Put u = tan x and v = tan y Diff u w.r.t. x & v w.r.t y 𝑑𝑒/𝑑π‘₯ = 𝑠𝑒𝑐2π‘₯ 𝑑𝑒/𝑠𝑒𝑐2π‘₯" " = dx 𝑑𝑣/𝑑𝑦 = 𝑠𝑒𝑐2𝑦 𝑑𝑒/𝑠𝑒𝑐2π‘₯" " = dy Therefore, our equation becomes ∫1β–’γ€–sec^2⁑π‘₯/tan⁑π‘₯ 𝑑π‘₯γ€—+∫1β–’γ€–sec^2⁑𝑦/tan⁑𝑦 𝑑𝑦=0γ€— ∫1β–’γ€–sec^2⁑π‘₯/tan⁑π‘₯ 𝑑𝑒/sec^2⁑π‘₯ γ€—+∫1β–’γ€– sec^2⁑𝑦/𝑣 𝑑𝑣/sec^2⁑𝑦 =0γ€— ∫1▒〖𝑑𝑒/𝑒+∫1▒〖𝑑𝑣/𝑣=0γ€—γ€— log |𝑒|+log⁑|𝑣|=log⁑𝑐 Putting back u = tan x and v = tan y log |tan⁑π‘₯ | + log |tan⁑𝑦 |=log⁑𝑐 log |tan⁑〖π‘₯+tan⁑𝑦 γ€— | =log⁑𝑐 tan x tan y = C (∡logβ‘γ€–π‘Ž+log⁑〖𝑏=logβ‘π‘Žπ‘ γ€— γ€— )

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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.