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Example 19 - Find general solution: dy/dx - y = cos x - Examples

Example 19 - Chapter 9 Class 12 Differential Equations - Part 2
Example 19 - Chapter 9 Class 12 Differential Equations - Part 3

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Example 19 Find the general solution of the differential equation 𝑑𝑦/𝑑π‘₯βˆ’π‘¦=cos⁑π‘₯ Differential equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = βˆ’1 & Q = cos x IF = e^∫1▒𝑝𝑑π‘₯ IF = e^(βˆ’βˆ«1β–’1𝑑π‘₯) IF = 𝑒^(βˆ’π‘₯) Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹) 𝑑π‘₯+𝑐〗 𝑦𝑒^(βˆ’π‘₯) = ∫1▒𝑒^(βˆ’π‘₯) cos⁑〖π‘₯+𝑐〗 Let I = ∫1▒𝑒^(βˆ’π‘₯) cos⁑〖π‘₯ 𝑑π‘₯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 = 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‘₯) sin⁑〖π‘₯ 𝑑π‘₯γ€— 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’ [sin⁑〖π‘₯ ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—βˆ’βˆ«1β–’γ€–(cos〗⁑〖π‘₯ ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—)γ€— "dx " γ€— ] I = βˆ’eβˆ’x cos x βˆ’ [βˆ’π‘’^(βˆ’π‘₯) sin⁑〖π‘₯ βˆ’βˆ«1β–’γ€–βˆ’π‘’^(βˆ’π‘₯) cos⁑π‘₯ 𝑑π‘₯γ€— " " 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 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.