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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 14 Find the general solution of the differential equation 𝑑𝑦/𝑑π‘₯βˆ’π‘¦=cos⁑π‘₯ Differential equation is of the form π’…π’š/𝒅𝒙+π‘·π’š=𝑸 where P = βˆ’1 & Q = cos x Finding Integrating Factor IF = e^∫1▒𝑝𝑑π‘₯ IF = e^(βˆ’βˆ«1β–’1𝑑π‘₯) IF = 𝒆^(βˆ’π’™) Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹) 𝑑π‘₯+𝑐〗 π’šπ’†^(βˆ’π’™) = ∫1▒𝒆^(βˆ’π’™) πœπ¨π¬β‘γ€–π’™+𝒄〗 Let I = ∫1▒𝒆^(βˆ’π’™) 𝒄𝒐𝒔⁑〖𝒙 𝒅𝒙〗 I = cos x ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—βˆ’ ∫1β–’[βˆ’sin⁑〖π‘₯∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—γ€— ]𝑑π‘₯ I = γ€–βˆ’π‘’γ€—^(βˆ’π‘₯)cos x βˆ’βˆ«1β–’γ€–βˆ’sin⁑〖π‘₯ (βˆ’π‘’^(βˆ’π‘₯) γ€—)γ€— 𝑑π‘₯ I = βˆ’eβˆ’x cos x βˆ’ ∫1▒〖𝒆^(βˆ’π’™) π’”π’Šπ’β‘γ€–π’™ 𝒅𝒙〗 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) = cos x & g (x) = 𝒆^"βˆ’x" I = βˆ’eβˆ’x cos x βˆ’ [sin⁑〖π‘₯ ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–(cos〗⁑〖π‘₯ ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—)γ€— "dx " 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) = sin x g (x) = eβˆ’x I = βˆ’eβˆ’x cos x βˆ’ [βˆ’π’†^(βˆ’π’™) π’”π’Šπ’β‘γ€–π’™ βˆ’βˆ«1β–’γ€–βˆ’π’†^(βˆ’π’™) 𝒄𝒐𝒔⁑𝒙 𝒅𝒙〗 " " γ€— ] I = βˆ’eβˆ’x cos x βˆ’ [βˆ’π‘’^(βˆ’π‘₯) sin⁑〖π‘₯+∫1▒〖𝑒^(βˆ’π‘₯) cos⁑π‘₯ 𝑑π‘₯γ€— " " γ€— ] I = βˆ’eβˆ’x cos x + 𝑒^(βˆ’π‘₯) sin⁑〖π‘₯βˆ’βˆ«1▒〖𝒆^(βˆ’π’™) 𝒄𝒐𝒔⁑𝒙 𝒅𝒙〗 " " γ€— I = eβˆ’x (sin x βˆ’ cos x) βˆ’ I 2I = eβˆ’x (sin x βˆ’ cos x) I = 𝒆^(βˆ’π’™)/𝟐 (sin x βˆ’ cos x) From (1) y 𝑒^(βˆ’π‘₯) = ∫1▒〖𝑒^(βˆ’π‘₯) cos⁑〖π‘₯+𝑐〗 γ€— y 𝑒^(βˆ’π‘₯) = 𝑒^(βˆ’π‘₯)/2 (sin x βˆ’ cos x) + c y = 𝟏/𝟐 (sin x βˆ’ cos x) + c𝒆^𝒙

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.