Ex 9.6, 10 - Find general solution: (x + y) dy/dx = 1 - Solving Linear differential equations - Equation given

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Ex 9.6, 10 For each of the differential equation given in Exercises 1 to 12, find the general solution : (π‘₯+𝑦) 𝑑𝑦/𝑑π‘₯=1 Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q or 𝑑π‘₯/𝑑𝑦 + P1x = Q1 (x + y) 𝑑𝑦/𝑑π‘₯ = 1 Dividing by (x + y), 𝑑𝑦/𝑑π‘₯ = 1/((π‘₯ + 𝑦)) 𝑑π‘₯/𝑑𝑦 = (π‘₯+𝑦) 𝑑π‘₯/𝑑𝑦 βˆ’ x = 𝑦 𝑑π‘₯/𝑑𝑦 + (βˆ’1) x = 𝑦 Step 2: Find P1 and Q1 Comparing (1) with 𝑑π‘₯/𝑑𝑦 + P1x = Q1 P1 = βˆ’1 and Q1 = y Step 3: Find Integrating factor, I.F. I.F. = 𝑒^∫1▒〖𝑃_1 𝑑𝑦〗 = e^∫1β–’γ€–(βˆ’1) 𝑑𝑦〗 = eβˆ’y So, I.F. = eβˆ’y Step 4 : Solution of the equation x Γ— I.F. = ∫1▒〖𝑄1×𝐼.𝐹. 𝑑𝑦+𝐢〗 Putting values, x Γ— e βˆ’ y = ∫1▒〖𝑦×𝑒^(βˆ’π‘¦).γ€— 𝑑𝑦+𝐢 Let I = ∫1▒〖𝑦.𝑒^(βˆ’π‘¦) 𝑑𝑦〗 = y ∫1▒〖𝑒^(βˆ’π‘¦) 𝑑𝑦〗 βˆ’ ∫1β–’[𝑑/𝑑𝑦 𝑦 ∫1▒〖𝑒^(βˆ’π‘¦) 𝑑𝑦〗] dy = y 𝑒^(βˆ’π‘¦)/(βˆ’1) βˆ’ ∫1β–’γ€–1. γ€— 𝑒^(βˆ’π‘¦)/(βˆ’1) dy. = βˆ’π‘¦.𝑒^(βˆ’π‘¦) + ∫1▒〖𝑒^(βˆ’π‘¦) 𝑑𝑦〗 = βˆ’π‘¦.𝑒^(βˆ’π‘¦) + 𝑒^(βˆ’π‘¦)/(βˆ’1) = βˆ’π‘¦.𝑒^(βˆ’π‘¦) – 𝑒^(βˆ’π‘¦) Putting value of I in (2) x e βˆ’y = ∫1▒〖𝑦×𝑒^(βˆ’π‘¦).γ€— 𝑑𝑦+𝐢 x e βˆ’y = βˆ’π‘¦π‘’^(βˆ’π‘¦)βˆ’π‘’^(βˆ’π‘¦)+𝐢 Dividing by 𝑒^(βˆ’π‘¦) x = βˆ’y βˆ’ 1 + Cey x + y + 1 = Cey

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