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

Transcript

Ex 9.6, 16 Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (π‘₯ , 𝑦) is equal to the sum of the coordinates of the point. We know that Slope of tangent to curve at (x, y) = 𝑑𝑦/𝑑π‘₯ Given that Slope of the tangent to the curve at any point (π‘₯ , 𝑦) is equal to the sum of the coordinates of the point. Therefore, 𝑑𝑦/𝑑π‘₯ = x + y 𝑑𝑦/𝑑π‘₯ βˆ’ y = x This is the form 𝑑𝑦/𝑑π‘₯+Py=Q where P = βˆ’1 & Q = x Finding Integrating factor IF = 𝑒^∫1▒〖𝑝𝑑π‘₯ γ€— IF = 𝑒^∫1β–’γ€–(βˆ’1)𝑑π‘₯ γ€— IF = eβˆ’x Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 𝑦𝑒^(βˆ’π‘₯) = ∫1β–’γ€–π‘₯𝑒^(βˆ’π‘₯) 𝑑π‘₯+𝑐〗 yeβˆ’x = x ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯βˆ’γ€— ∫1β–’γ€–[1∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—] 𝑑π‘₯+𝑐〗 yeβˆ’x = βˆ’x 𝑒^(βˆ’π‘₯) βˆ’βˆ«1β–’γ€–βˆ’π‘’^(βˆ’π‘₯) 𝑑π‘₯+𝑐〗 yeβˆ’x = βˆ’x 𝑒^(βˆ’π‘₯)+∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯+𝑐〗 yeβˆ’x = βˆ’x 𝑒^(βˆ’π‘₯)+(βˆ’π‘’^(βˆ’π‘₯))/(βˆ’1) +𝑐 yeβˆ’x = βˆ’x 𝑒^(βˆ’π‘₯)βˆ’π‘’^(βˆ’π‘₯) +𝑐 Dividing both sides by eβˆ’x y = βˆ’x βˆ’ 1 + 𝑐/𝑒^(βˆ’π‘₯) 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) = x & g (x) = 𝑒^(βˆ’π‘₯) y = βˆ’x βˆ’ 1 + cex Since curve passes through origin, Putting x = 0 & y = 0 in (2) 0 = 0 βˆ’ 1 + Ce0 1 = C C = 1 Put value of C in (1) y = βˆ’x βˆ’ 1 + ex x + y + 1 = ex …(1)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.