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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Ex 9.4, 15 Find the equation of curve passing through the point (0 , 0) and whose differential equation is 𝑦^β€²=𝑒^π‘₯ sin⁑π‘₯ 𝑦^β€² = 𝑒^π‘₯ sin x 𝑑𝑦/𝑑π‘₯ = 𝑒^π‘₯ sin x dx 𝑑𝑦 = 𝑒^π‘₯ sin x dx Integrating both sides ∫1▒𝑑𝑦 = ∫1▒〖𝑒π‘₯ sin⁑〖π‘₯ 𝑑π‘₯γ€— γ€— y = ∫1▒〖𝑒π‘₯ sin⁑〖π‘₯ 𝑑π‘₯γ€— γ€— Integrating by parts, using formula ∫1▒〖𝑓 (π‘₯)𝑔(π‘₯)𝑑π‘₯ γ€—= f(π‘₯) ∫1▒〖𝑔(π‘₯)𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓′(π‘₯)∫1▒𝑔(π‘₯)𝑑π‘₯] 𝑑π‘₯ γ€— γ€— take f(x) = ex and g (x) = sin x y = 𝑒^π‘₯ ∫1▒〖〖𝑠𝑖𝑛 π‘₯〗⁑〖 𝑑π‘₯γ€—βˆ’γ€— ∫1β–’[(𝑒^π‘₯) ∫1β–’γ€–sin⁑π‘₯ 𝑑π‘₯ γ€—]𝑑π‘₯ y = 𝑒^π‘₯ (βˆ’cos⁑π‘₯) βˆ’ ∫1β–’γ€–(βˆ’cos⁑π‘₯)γ€— 𝑒^π‘₯ 𝑑π‘₯ y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+∫1▒〖𝑒^π‘₯ cos⁑π‘₯ 𝑑π‘₯γ€— Integrating by parts, using formula ∫1▒〖𝑓 (π‘₯)𝑔(π‘₯)𝑑π‘₯ γ€—= f(π‘₯) ∫1▒〖𝑔(π‘₯)𝑑π‘₯ βˆ’βˆ«1β–’γ€–[𝑓′(π‘₯)∫1▒𝑔(π‘₯)𝑑π‘₯] 𝑑π‘₯ γ€— γ€— take f(x) = ex and g(x) = cos x y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+ 𝑒^π‘₯ ∫1β–’γ€–γ€–π‘π‘œπ‘  π‘₯〗⁑〖 𝑑π‘₯γ€—βˆ’γ€— ∫1β–’[(𝑒^π‘₯) ∫1β–’γ€–cos⁑π‘₯ 𝑑π‘₯ γ€—]𝑑π‘₯ y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+𝑒^π‘₯ 𝑠𝑖𝑛⁑π‘₯ βˆ’βˆ«1▒〖𝑒^π‘₯ sin⁑π‘₯ 𝑑π‘₯γ€— But From (1) y = ∫1▒〖𝑒π‘₯ 𝑠𝑖𝑛⁑〖π‘₯ 𝑑π‘₯γ€— γ€— y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+𝑒^π‘₯ 𝑠𝑖𝑛⁑π‘₯ βˆ’π‘¦ y + y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+𝑒^π‘₯ 𝑠𝑖𝑛⁑π‘₯ 2y = 𝑒^π‘₯ (sin⁑π‘₯βˆ’cos⁑π‘₯) y = 1/2 𝑒^π‘₯ (sin⁑π‘₯βˆ’cos⁑π‘₯ )+ C Given curve passes through (0, 0) Putting x = 0, y = 0 in equation 0 = 1/2 𝑒^0 (sin⁑0βˆ’cos⁑0) + C 0 = 1/2 (0 βˆ’1) + C 0 = (βˆ’1)/2 + C C = 1/2 Putting value of C in (1) y = 1/2 𝑒^π‘₯ (sin⁑π‘₯βˆ’cos⁑π‘₯ ) + 1/2 y – 1/2= 1/2 𝑒^π‘₯ (sin⁑π‘₯βˆ’cos⁑π‘₯ ) (2𝑦 βˆ’ 1)/2= 1/2(𝑒^π‘₯ sin⁑π‘₯+𝑒^π‘₯ cos⁑π‘₯) 2y – 1 = (𝒆^𝒙 π’”π’Šπ’β‘π’™+𝒆^𝒙 𝒄𝒐𝒔⁑𝒙)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.