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Ex 9.4, 10 - Find general solution: ex tan y dx + (1 - ex) - Variable separation - Equation given

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.4, 10 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯+(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦=0γ€— γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯+(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦=0γ€— γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯=βˆ’(1βˆ’π‘’^π‘₯ ) sec^2⁑〖𝑦 𝑑𝑦〗 γ€— 𝑒^π‘₯ tan⁑〖𝑦 𝑑π‘₯γ€—=(𝑒^π‘₯βˆ’1) sec^2⁑〖𝑦 𝑑𝑦〗 𝑒^π‘₯/(𝑒^π‘₯ βˆ’ 1) dx = (𝑠𝑒𝑐2𝑦 𝑑𝑦)/tan⁑𝑦 Integrating both sides. ∫1▒〖𝑒^π‘₯/(𝑒^π‘₯ βˆ’ 1) 𝑑π‘₯γ€— = ∫1β–’γ€–(𝑠𝑒𝑐2 𝑦)/tan⁑𝑦 𝑑𝑦〗 Put 𝑒^π‘₯βˆ’1 = u and put tan y = v Diff u w.r.t. x & v w.r.t y Therefore ∫1▒〖𝑒π‘₯/𝑒 𝑑𝑒/𝑒π‘₯γ€— = ∫1▒𝑠𝑒𝑐2𝑦/(𝑣 𝑠𝑒𝑐2𝑦) dv ∫1▒𝑑𝑒/𝑒 = ∫1▒𝑑𝑣/𝑣 log u + c1 = log v Putting back u = ex βˆ’ 1 and V = tan y log |"ex βˆ’ 1" | + c1 = log tan y Putting c1 = log c log |"ex βˆ’ 1" |+ log c = log (tan y) log |𝑐("ex βˆ’ 1" )|= log |tan⁑𝑦 | 𝑐("ex βˆ’ 1" ) = tan y tan y = c ("ex βˆ’ 1" ) is the general solution

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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