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Transcript

Ex 9.5, 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 π’…π’š/𝒅𝒙 + (𝟏/𝒙+𝒄𝒐𝒕⁑𝒙 ) y = 1 Comparing (1) with 𝑑𝑦/𝑑π‘₯ + Py = Q P = 𝟏/𝒙 + 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 Solution of the equation is y Γ— I.F. = ∫1β–’γ€–Q×𝐼𝐹〗⁑𝑑π‘₯ + C y (x sin x) = ∫1▒〖𝒙.γ€–π’”π’Šπ’ 𝒙〗⁑𝒅𝒙 γ€— Let I = ∫1▒〖𝒙.𝐬𝐒𝐧⁑〖𝒙.𝒅𝒙〗 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€— 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⁑𝒙)/(𝒙 π’”π’Šπ’β‘π’™ ) + π’”π’Šπ’β‘π’™/(𝒙 π’”π’Šπ’β‘π’™ ) + π‘ͺ/(𝒙 π’”π’Šπ’β‘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 = 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⁑𝒙)/(𝒙 π’”π’Šπ’β‘π’™ ) + π’”π’Šπ’β‘π’™/(𝒙 π’”π’Šπ’β‘π’™ ) + π‘ͺ/(𝒙 π’”π’Šπ’β‘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 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.