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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.6, 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‘₯ ) 𝑑𝑦/𝑑π‘₯ + (1/(π‘₯ log⁑π‘₯ ))𝑦=2/π‘₯^2 Step 2: Find P and Q ...(1) Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 1/(π‘₯ log⁑π‘₯ ) & Q = 2/π‘₯2 Step 3: Find Integration factor, I.F IF = e^∫1▒〖𝑝 𝑑π‘₯γ€— IF = e^∫1β–’γ€–1/(π‘₯ log⁑π‘₯ ) 𝑑π‘₯γ€— 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 = |𝑑| 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–’2/π‘₯2 . log x. dx + C (As log x is always positive) ...(2) Let I = 2 ∫1β–’log⁑〖π‘₯ π‘₯^(βˆ’2) 𝑑π‘₯γ€— Solving I I = 2 ∫1β–’log⁑〖π‘₯ π‘₯^(βˆ’2) 𝑑π‘₯γ€— I = 2["log x. " ∫1β–’γ€–π‘₯^(βˆ’2) 𝑑π‘₯βˆ’βˆ«1β–’ 1/π‘₯ [∫1β–’γ€– π‘₯^(βˆ’2) 𝑑π‘₯γ€—] γ€— 𝑑π‘₯" " ] 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€—] 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) = log x & g (x) = π‘₯^(βˆ’2) = 2[(βˆ’1)/π‘₯ " .log x βˆ’ " 1/π‘₯] = (βˆ’2)/π‘₯ (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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.