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

Transcript

Ex 9.6, 1 For each of the differential equation given in Exercises 1 to 12, find the general solution : 𝑑𝑦/𝑑π‘₯+2𝑦=𝑠𝑖𝑛π‘₯ Step 1: Put in form 𝑑𝑦/𝑑π‘₯ + Py = Q 𝑑𝑦/𝑑π‘₯+2𝑦=sin⁑π‘₯ Step 2: Find P and Q Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q ∴ P = 2 and Q = sin x ...(1) Step 3: Find integrating factor, IF IF = e^∫1▒𝑃𝑑π‘₯ IF = 𝑒^∫1β–’2𝑑π‘₯= 𝑒^2π‘₯ So, IF = 𝑒^2π‘₯ Step 4 : Solution of the equation y Γ— I.F = ∫1▒〖𝑄×𝐼.𝐹.𝑑π‘₯+𝑐 γ€— Putting values, 𝑦×𝑒^2π‘₯ = ∫1β–’γ€–sin⁑π‘₯ 𝑒^2π‘₯ 𝑑π‘₯γ€—+𝑐 𝐿𝑒𝑑 𝐼= ∫1β–’γ€–sin⁑π‘₯ 𝑒^2π‘₯ 𝑑π‘₯γ€— ...(2) Solving I 𝐼= ∫1β–’γ€–sin⁑〖π‘₯.𝑒^2π‘₯ γ€—.𝑑π‘₯ γ€— = sin x ∫1▒〖𝑒^2π‘₯.𝑑π‘₯βˆ’βˆ«1β–’γ€–[𝑑/𝑑π‘₯ sin⁑π‘₯ ∫1▒〖𝑒^2π‘₯ 𝑑π‘₯ γ€—] γ€— γ€— = sin x 𝑒^2π‘₯/2 βˆ’ ∫1β–’cos⁑π‘₯ 𝑒^2π‘₯/2 dx = 1/2 sin⁑〖π‘₯ 𝑒^2π‘₯ γ€—βˆ’1/2 [cos⁑π‘₯ ∫1▒𝑒^2π‘₯ 𝑑π‘₯ βˆ’βˆ«1β–’β–ˆ(𝑑@𝑑π‘₯) cos⁑π‘₯ ∫1▒𝑒^2π‘₯ 𝑑π‘₯ ]dx Again using 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) = cos x & g(x) = 𝑒^2π‘₯ = 1/2 sin⁑〖π‘₯ 𝑒^2π‘₯ γ€—βˆ’1/2 [cos⁑π‘₯ ∫1▒𝑒^2π‘₯/2 βˆ’βˆ«1β–’γ€–(βˆ’sin x)γ€— ∫1▒𝑒^2π‘₯/2 𝑑π‘₯ ] = 1/2 sin⁑〖π‘₯ 𝑒^2π‘₯ γ€—βˆ’1/2 [cos⁑π‘₯ 𝑒^2π‘₯/2+1/2 ∫1▒〖𝐬𝐒𝐧 𝐱 𝒆^πŸπ’™ 𝒅𝒙〗] = 1/2 sin⁑〖π‘₯ 𝑒^2π‘₯ γ€—βˆ’1/2 [(cos⁑π‘₯ 𝑒^2π‘₯)/2+1/2 𝑰] + C I = 1/2 sin x 𝑒^2π‘₯ βˆ’1/4 cos x 𝑒^2π‘₯ βˆ’ 1/4 I + C I + 1/4 I = 1/4 [2 sin⁑〖π‘₯ 𝑒^2π‘₯ βˆ’cos⁑〖π‘₯ 𝑒^2π‘₯ γ€— γ€— ] + C 5𝐼/4 = 𝑒^2π‘₯/4 [2 sin⁑〖π‘₯βˆ’cos⁑π‘₯ γ€— ] + C 𝐼 = 𝑒^2π‘₯/5 [2 sin⁑〖π‘₯βˆ’cos⁑π‘₯ γ€— ] + C (From 3) Now, Putting value of I in (2) y 𝑒^2π‘₯ = 𝑒^2π‘₯/5 [2 sin⁑〖π‘₯ βˆ’cos⁑π‘₯ γ€— ] + C Dividing by e2x y = 𝟏/πŸ“ [𝟐 𝐬𝐒𝐧⁑〖𝒙 βˆ’πœπ¨π¬β‘π’™ γ€— ]+π‘ͺ𝒆^(βˆ’πŸπ’™)

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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.