Misc 11 - Find particular solution: (x - y) (dx + dy) = dx - dy - Solving homogeneous differential equation

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Misc 11 Find a particular solution of the differential equation (π‘₯βˆ’π‘¦)(𝑑π‘₯+𝑑𝑦)=𝑑π‘₯βˆ’π‘‘π‘¦ , given that 𝑦=βˆ’1 , when π‘₯=0 (𝐻𝑖𝑛𝑑:𝑝𝑒𝑑 π‘₯βˆ’π‘¦=𝑑) (π‘₯βˆ’π‘¦)(𝑑π‘₯+𝑑𝑦)=𝑑π‘₯βˆ’π‘‘π‘¦ π‘₯𝑑π‘₯ + π‘₯𝑑𝑦 βˆ’ y dx βˆ’ y dy = dx βˆ’ dy x dx βˆ’ y dx βˆ’ dx = βˆ’ xdy + y dy βˆ’ dy (x βˆ’ y βˆ’ 1) dx = (βˆ’x + y βˆ’ 1)dy 𝑑𝑦/𝑑π‘₯ = (π‘₯ βˆ’ 𝑦 βˆ’ 1)/(βˆ’π‘₯ + 𝑦 βˆ’ 1) 𝑑𝑦/𝑑π‘₯ = ((π‘₯ βˆ’ 𝑦 βˆ’ 1))/(βˆ’(π‘₯ βˆ’π‘¦ + 1)) Put x βˆ’ y = t Diff w.r.t.x 1 βˆ’ 𝑑𝑦/𝑑π‘₯ = 𝑑𝑑/𝑑π‘₯ Putting value of t & dt in (1) 1 βˆ’ 𝑑𝑑/𝑑π‘₯ = βˆ’ ((𝑑 βˆ’ 1))/(𝑑 + 1) 𝑑𝑑/𝑑π‘₯ = 1 + ((𝑑 βˆ’ 1))/(𝑑 + 1) 𝑑𝑑/𝑑π‘₯ = (𝑑 + 1 + 𝑑 βˆ’ 1)/(𝑑 + 1) 𝑑𝑑/𝑑π‘₯ = 2𝑑/(𝑑 βˆ’ 1) (𝑑 βˆ’ 1)/2𝑑 dt = 𝑑π‘₯ Integrating both sides ∫1β–’γ€–(𝑑 + 1 )/2𝑑 𝑑𝑑〗 = ∫1▒𝑑π‘₯ ∫1β–’γ€–(𝑑/2𝑑+1/2𝑑) 𝑑𝑑〗 = x + c ∫1β–’(1/2+1/2𝑑)𝑑𝑑 = x + c 𝑑/2 + 1/2 log |𝑑| = x + c Putting value of t = x – y (π‘₯ βˆ’ 𝑦)/2 + 1/2 log |π‘₯βˆ’π‘¦| = x + C Given y = –1 when x = 0 Put x = 0 & y = βˆ’1 in (2) (0 + 1)/2 + 1/2 log 1 = 0 + C 1/2 = C C = 1/2 Putting value in (2) (π‘₯ βˆ’ 𝑦)/2+1/2 log⁑〖|π‘₯ βˆ’ 𝑦|=π‘₯+1/2γ€— (π‘₯ βˆ’ 𝑦)/2+1/2 log⁑〖|π‘₯ βˆ’ 𝑦|=(2π‘₯ + 1)/2γ€— x βˆ’ y + log |π‘₯βˆ’π‘¦|=2π‘₯+1 log |π‘₯βˆ’π‘¦| = 2x + 1 βˆ’ x + y log |π’™βˆ’π’š| = x + y + 1

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Davneet Singh
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.