Slide1.JPG

Slide2.JPG
Slide3.JPG

Go Ad-free

Transcript

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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 and Computer Science at Teachoo