Check sibling questions

Slide23.JPG

Slide24.JPG
Slide25.JPG

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


Transcript

Ex 9.3, 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^𝒙/(𝒆^𝒙 βˆ’ 𝟏) dx = (π’”π’†π’„πŸπ’š π’…π’š)/π’•π’‚π’β‘π’š Integrating both sides. ∫1▒〖𝑒^π‘₯/(𝑒^π‘₯ βˆ’ 1) 𝑑π‘₯γ€— = ∫1β–’γ€–(𝑠𝑒𝑐2 𝑦)/tan⁑𝑦 𝑑𝑦〗 Put 𝒆^π’™βˆ’πŸ = u and put tan y = v Diff u w.r.t. x ex = 𝑑𝑒/𝑑π‘₯ dx = 𝒅𝒖/𝒆𝒙 Diff v w.r.t. y sec2 y = 𝑑𝑣/𝑑𝑦 dy = 𝒅𝒗/γ€–π¬πžπœγ€—^πŸβ‘π’š 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

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.