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Transcript

Ex 9.5, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : π‘₯π‘™π‘œπ‘”π‘₯ 𝑑𝑦/𝑑π‘₯+𝑦=2/π‘₯ π‘™π‘œπ‘”π‘₯ Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q xlog x 𝑑𝑦/𝑑π‘₯ + y = 2/π‘₯ log x Dividing by x log x, 𝑑𝑦/𝑑π‘₯+𝑦" Γ— " 1/(π‘₯ log⁑π‘₯ ) = 2/π‘₯ π‘™π‘œπ‘” π‘₯" Γ— " 1/(π‘₯ log⁑π‘₯ ) π’…π’š/𝒅𝒙 + (𝟏/(𝒙 π’π’π’ˆβ‘π’™ ))π’š=𝟐/𝒙^𝟐 Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 𝟏/(𝒙 π’π’π’ˆβ‘π’™ ) & Q = 𝟐/π’™πŸ Step 3: Find Integration factor, I.F IF = e^∫1▒〖𝑝 𝑑π‘₯γ€— IF = 𝐞^∫1β–’γ€–πŸ/(𝒙 π₯𝐨𝐠⁑𝒙 ) 𝒅𝒙〗 Let t = log x dt = 1/π‘₯ dx dx = x dt So, IF = e^∫1β–’γ€–1/(π‘₯ 𝑑) Γ— π‘₯𝑑𝑑〗 IF = e^∫1β–’γ€–1/𝑑 𝑑𝑑〗 IF = e^log⁑〖|𝑑|γ€— IF = |𝒕| Putting back t = log x IF = |log x| IF = log x Step 4: Solution of the equation y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹. 𝑑π‘₯+𝐢〗 Putting values, y Γ— log x = ∫1β–’πŸ/π’™πŸ . log x. dx + C Let I = 2 ∫1β–’π’π’π’ˆβ‘γ€–π’™ 𝒙^(βˆ’πŸ) 𝒅𝒙〗 Solving I I = 2 ∫1β–’π’π’π’ˆβ‘γ€–π’™ 𝒙^(βˆ’πŸ) 𝒅𝒙〗 I = 2["log x. " ∫1▒〖𝒙^(βˆ’πŸ) π’…π’™βˆ’βˆ«1β–’ 𝟏/𝒙 [∫1β–’γ€– 𝒙^(βˆ’πŸ) 𝒅𝒙〗] γ€— 𝒅𝒙" " ] I = 2 ["log x . " π‘₯^(βˆ’1)/((βˆ’1)) " βˆ’ " ∫1β–’γ€– 1/π‘₯γ€— " . " ((π‘₯^(βˆ’1)))/((βˆ’1)) ".dx " ] = 2["βˆ’ log x. " 1/π‘₯ " + " ∫1β–’γ€–1/π‘₯^2 .𝑑π‘₯γ€—] = 2[(βˆ’1)/π‘₯ " .log x βˆ’ " 1/π‘₯] = (βˆ’πŸ)/𝒙 (1 + log x) Putting value of I in (2) y log x = I + C y. log x = (βˆ’πŸ)/𝒙 (1 + log x) + C Which is the general solution of the given equation.

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.