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


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Ex 9.5, 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 𝒅𝒙/π’…π’š + (βˆ’1) x = π’š 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 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 Γ— 𝑒^(βˆ’π‘¦).γ€— 𝑑𝑦+𝐢 Let I = ∫1β–’γ€–π’š.𝒆^(βˆ’π’š) π’…π’šγ€— Using Integration by parts ∫1▒〖𝑓(𝑦)𝑔(𝑦)𝑑𝑦=𝑓(𝑦) ∫1▒〖𝑔(𝑦)π‘‘π‘¦βˆ’γ€—γ€— ∫1β–’[𝑓′(𝑦)∫1▒𝑔(𝑦)𝑑𝑦] 𝑑𝑦 Taking 𝑓(𝑦) = y and g (y) = 𝑒^(βˆ’π‘¦) = y ∫1▒〖𝒆^(βˆ’π’š) π’…π’šγ€— βˆ’ ∫1β–’[𝒅/π’…π’š π’š ∫1▒〖𝒆^(βˆ’π’š) π’…π’šγ€—] 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’šπ’†^(βˆ’π’š)βˆ’π’†^(βˆ’π’š)+π‘ͺ Dividing by 𝑒^(βˆ’π‘¦) x = βˆ’y βˆ’ 1 + Cey x + y + 1 = Cey

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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.