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Ex 9.6, 9 - Find general solution: x dy/dx + y - x + xy cot x

Ex 9.6, 9 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.6, 9 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.6, 9 - Chapter 9 Class 12 Differential Equations - Part 4


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.