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

Transcript

Ex 9.6, 9 For each of the differential equation find the general solution : π‘₯ 𝑑𝑦/𝑑π‘₯+π‘¦βˆ’π‘₯+π‘₯𝑦 cot⁑〖π‘₯=0(π‘₯β‰ 0)γ€— Given equation x 𝑑𝑦/𝑑π‘₯ + y βˆ’ x + xy cot x = 0 Dividing both sides by x 𝑑𝑦/𝑑π‘₯ + 𝑦/π‘₯ βˆ’ 1 + y cot x = 0 𝑑𝑦/𝑑π‘₯ + y (1/π‘₯+cot⁑π‘₯ ) βˆ’ 1 = 0 𝑑𝑦/𝑑π‘₯ + (1/π‘₯+cot⁑π‘₯ ) y = 1 …(1) Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 1/π‘₯ + cot x & Q = 1 Finding integrating factor, I.F. I.F. = e^∫1▒〖𝑝 𝑑π‘₯ γ€— = e^∫1β–’(1/π‘₯ + cot⁑π‘₯ )𝑑π‘₯ = e^∫1β–’γ€–1/π‘₯ 𝑑π‘₯ + ∫1β–’γ€–cot⁑π‘₯ 𝑑π‘₯γ€—γ€— = 𝑒^(log⁑π‘₯ + log⁑sin⁑π‘₯ ) = 𝑒^log⁑〖(π‘₯ sin⁑π‘₯)γ€— = x sin x So, I.F. = x sin x (Using log a + log b = log ab) (Using 𝑒^log⁑π‘₯ = x) Solution of the equation is y Γ— I.F. = ∫1β–’γ€–Q×𝐼𝐹〗⁑𝑑π‘₯ + C y (x sin x) = ∫1β–’γ€–π‘₯.〖𝑠𝑖𝑛 π‘₯〗⁑𝑑π‘₯ γ€— Let I = ∫1β–’γ€–π‘₯.sin⁑〖π‘₯.𝑑π‘₯γ€— γ€— I = x ∫1β–’sin⁑〖π‘₯ 𝑑π‘₯βˆ’βˆ«1β–’[1.∫1β–’sin⁑〖π‘₯ 𝑑π‘₯γ€— ]𝑑π‘₯γ€— = x (βˆ’ cos x) βˆ’ ∫1β–’γ€–1.(βˆ’cos⁑〖π‘₯)γ€— 𝑑π‘₯γ€— = βˆ’ x. cos x + ∫1β–’cos⁑〖π‘₯ 𝑑π‘₯γ€— …(2) Using formula ∫1▒〖𝑓(π‘₯)𝑔(π‘₯)𝑑π‘₯=𝑓(π‘₯)𝑓𝑔(π‘₯)𝑑π‘₯βˆ’βˆ«1β–’[𝑓′(π‘₯)][𝑔(π‘₯)𝑑π‘₯] γ€— dx Taking f(x) = x & g(x) = sin x = βˆ’ x cos x + sin x Putting value of I in (2), y x sin x = βˆ’x cos x + sin x + C Divide by x sin x y = (βˆ’π‘₯ cos⁑π‘₯)/(π‘₯ sin⁑π‘₯ ) + sin⁑π‘₯/(π‘₯ sin⁑π‘₯ ) + 𝐢/(π‘₯ sin⁑π‘₯ ) y = βˆ’cot x + 1/π‘₯ + 𝐢/(π‘₯ 𝑠𝑖𝑛⁑π‘₯ ) y = 𝟏/𝒙 βˆ’ cot x + π‘ͺ/(𝒙 π’”π’Šπ’β‘π’™ ) Which is the general solution of the given differential 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.