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Ex 9.6, 16 - Find equation of curve passing through origin given that

Ex 9.6, 16 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.6, 16 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.6, 16 - Chapter 9 Class 12 Differential Equations - Part 4

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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 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.