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Ex 9.5, 6 For each of the differential equation given in Exercises 1 to 12, find the general solution : π‘₯ 𝑑𝑦/𝑑π‘₯+2𝑦=π‘₯^2 π‘™π‘œπ‘”π‘₯ Step 1 : Convert into 𝑑𝑦/𝑑π‘₯ + py = Q π‘₯ 𝑑𝑦/𝑑π‘₯+2𝑦=π‘₯^2 π‘™π‘œπ‘”π‘₯ Dividing both sides by x π’…π’š/𝒅𝒙 + πŸπ’š/𝒙 = x log x Step 2 : Find P and Q Differential equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = 𝟐/𝒙 and Q = x log x Step 3 : Finding integrating factor IF = 𝑒^∫1▒〖𝑝 𝑑π‘₯γ€—epdx IF = 𝑒^∫1β–’γ€–2/π‘₯ 𝑑π‘₯γ€— IF = 𝑒^(2∫1β–’γ€–1/π‘₯ 𝑑π‘₯γ€—) IF = 𝑒^(2 log⁑π‘₯ ) IF = 𝑒^log⁑〖π‘₯^2 γ€— IF = 𝒙^𝟐 Step 4 : Solution of the equation Solution is y (IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 yx2 = ∫1β–’γ€–π‘₯ log⁑〖π‘₯Γ—π‘₯^2 𝑑π‘₯+𝑐〗 γ€— yx2 = ∫1β–’γ€–π’π’π’ˆβ‘π’™ 𝒙^πŸ‘ γ€— + 𝒄 Integrating by parts with ∫1▒〖𝑓(π‘₯) 𝑔(π‘₯) 𝑑π‘₯=𝑓(π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓^β€² (π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯] 𝑑π‘₯γ€—γ€—γ€—γ€— Take f (x) = sin x & g (x) = 𝑒^2π‘₯ yx2 = log x ∫1β–’γ€–π‘₯^3 𝑑π‘₯βˆ’βˆ«1β–’[𝑑/𝑑π‘₯ log⁑〖π‘₯ γ€— ∫1β–’γ€–π‘₯^3 𝑑π‘₯γ€—] γ€—dx yx2 = log x (π‘₯^4/4)βˆ’βˆ«1β–’1/π‘₯ (π‘₯^4/4)𝑑π‘₯+𝑐 yx2 = (π‘₯^4 log⁑π‘₯)/4 βˆ’ ∫1β–’π‘₯^3/4 𝑑π‘₯+𝑐 yx2 = (π‘₯^4 log⁑π‘₯)/4 βˆ’ π‘₯^4/(4 Γ— 4)+𝑐 yx2 = (𝒙^πŸ’ π’π’π’ˆβ‘π’™)/πŸ’ βˆ’ 𝒙^πŸ’/πŸπŸ”+𝒄 y = (π‘₯^4 log⁑π‘₯)/(4π‘₯^2 ) βˆ’ π‘₯^4/(16π‘₯^2 ) + 𝐢/π‘₯^2 y = (π‘₯^2 log⁑〖|π‘₯|γ€—)/4 βˆ’ π‘₯^2/16 + 𝑐π‘₯^(βˆ’2) y = 𝒙^𝟐/πŸπŸ” (4 log |"x" | βˆ’ 1) + 𝒄𝒙^(βˆ’πŸ)

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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.