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Ex 9.3, 15 Find the equation of curve passing through the point (0 , 0) and whose differential equation is 𝑦^β€²=𝑒^π‘₯ sin⁑π‘₯ 𝑑𝑦/𝑑π‘₯ = 𝑒^π‘₯ sin x 𝑑𝑦 = 𝑒^π‘₯ sin x dx Integrating both sides ∫1▒𝑑𝑦 = ∫1▒〖𝑒π‘₯ sin⁑〖π‘₯ 𝑑π‘₯γ€— γ€— y = ∫1▒〖𝒆𝒙 π’”π’Šπ’β‘γ€–π’™ 𝒅𝒙〗 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€— Taking f(x) = ex and g (x) = sin x 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^𝒙 𝒄𝒐𝒔⁑𝒙 𝒅𝒙〗 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€— Taking 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^𝒙 𝐬𝐒𝐧⁑𝒙 𝒅𝒙〗 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’š y + y = –𝑒^π‘₯ π‘π‘œπ‘ β‘π‘₯+𝑒^π‘₯ 𝑠𝑖𝑛⁑π‘₯ 2y = 𝑒^π‘₯ (sin⁑π‘₯βˆ’cos⁑π‘₯) y = 𝑒^π‘₯ ∫1▒〖〖𝑠𝑖𝑛 π‘₯〗⁑〖 𝑑π‘₯γ€—βˆ’γ€— ∫1β–’[(𝑒^π‘₯) ∫1β–’γ€–sin⁑π‘₯ 𝑑π‘₯ γ€—]𝑑π‘₯ y = 𝟏/𝟐 𝒆^𝒙 (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘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 (0βˆ’1) + C 0 = (βˆ’1)/2 + C C = 𝟏/𝟐 Putting value of C in (1) y = 𝟏/𝟐 𝒆^𝒙 (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘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‘₯βˆ’cos⁑π‘₯ ) 2y – 1 = 𝒆^𝒙 (π’”π’Šπ’β‘π’™βˆ’π’„π’π’”β‘π’™ )

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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 and Computer Science at Teachoo