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

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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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‘₯ tan⁑𝑦 𝑑π‘₯ +sec^2⁑𝑦 tan⁑〖π‘₯ 𝑑𝑦〗)/tan⁑〖𝑦 tan⁑π‘₯ γ€— = 0/tan⁑〖π‘₯ tan⁑𝑦 γ€— (𝑠𝑒𝑐2π‘₯ π‘‘π‘Žπ‘›π‘¦ 𝑑π‘₯)/tan⁑〖𝑦 tan⁑π‘₯ γ€— + (𝑠𝑒𝑐2𝑦 π‘‘π‘Žπ‘›π‘₯ 𝑑𝑦)/tan⁑〖𝑦 tan⁑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 Therefore ∫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

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.