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


Ex 9.3, 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 (π’”π’†π’„πŸπ’™ )/π­πšπ§β‘π’™ dx + π’”π’†π’„πŸπ’š/π’•π’‚π’β‘π’š dy = 0 Integrating both sides ∫1β–’γ€–(𝑠𝑒𝑐2π‘₯/tan⁑π‘₯ 𝑑π‘₯+𝑠𝑒𝑐2𝑦/tan⁑𝑦 𝑑𝑦)=γ€— 0 ∫1β–’γ€–π’”π’†π’„πŸπ’™/𝒕𝒂𝒏⁑𝒙 𝒅𝒙+∫1β–’γ€–π’”π’†π’„πŸπ’š/π’•π’‚π’β‘π’š π’…π’šγ€—=γ€— 0 Put u = tan x and v = tan y Diff u w.r.t. x & v w.r.t y Therefore, our equation becomes ∫1β–’γ€–sec^2⁑π‘₯/tan⁑π‘₯ 𝑑π‘₯γ€—+∫1β–’γ€–sec^2⁑𝑦/tan⁑𝑦 𝑑𝑦=0γ€— 𝑑𝑒/𝑑π‘₯ = 𝑠𝑒𝑐2π‘₯ 𝒅𝒖/π’”π’†π’„πŸπ’™" " = dx 𝑑𝑣/𝑑𝑦 = 𝑠𝑒𝑐2𝑦 𝒅𝒖/π’”π’†π’„πŸπ’™" " = dy ∫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 |𝒖|+π’π’π’ˆβ‘|𝒗|=π’π’π’ˆβ‘π’„ Putting back u = tan x and v = tan y log |𝒕𝒂𝒏⁑𝒙 | + log |π­πšπ§β‘π’š |=π₯𝐨𝐠⁑𝒄 log |tan⁑〖π‘₯+tan⁑𝑦 γ€— | =log⁑𝑐 tan x tan y = C

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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.