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Ex 9.6, 17 - Curve through (0,2), sum of coordinate of point

Ex 9.6, 17 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.6, 17 - Chapter 9 Class 12 Differential Equations - Part 3
Ex 9.6, 17 - Chapter 9 Class 12 Differential Equations - Part 4
Ex 9.6, 17 - Chapter 9 Class 12 Differential Equations - Part 5

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Transcript

Ex 9.6, 17 Find the equation of a curve passing through the point(0 , 2) given that the sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 We know that Slope of tangent to curve at (x, y) = 𝑑𝑦/𝑑π‘₯ Given that sum of the coordinate of any point of curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5 Therefore, |𝑑𝑦/𝑑π‘₯| + 5 = x + y |𝑑𝑦/𝑑π‘₯| = x + y – 5 𝑑𝑦/𝑑π‘₯ = Β± (x + y βˆ’ 5) So, we will take both positive sign and negative sign and then solve it Taking (+) ve sign 𝑑𝑦/𝑑π‘₯ = x + y βˆ’ 5 𝑑𝑦/𝑑π‘₯ βˆ’ y = x βˆ’ 5 Equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = βˆ’1 & Q = x βˆ’ 5 IF = e^∫1▒𝑃𝑑π‘₯ IF = e^(∫1β–’γ€–(βˆ’1)γ€— 𝑑π‘₯) IF = 𝑒^(βˆ’π‘₯) Taking (βˆ’) ve sign 𝑑𝑦/𝑑π‘₯ = βˆ’x βˆ’ y + 5 𝑑𝑦/𝑑π‘₯ + y = βˆ’x + 5 Equation is of the form 𝑑𝑦/𝑑π‘₯+𝑃𝑦=𝑄 where P = 1 & Q = βˆ’x + 5 IF = e^∫1▒𝑃𝑑π‘₯ IF = e^∫1β–’1𝑑π‘₯ IF = e^π‘₯ Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 yeβˆ’x = ∫1β–’γ€–(π‘₯βˆ’5) 𝑒^(βˆ’π‘₯) 𝑑π‘₯+𝑐〗 yeβˆ’x = (x βˆ’ 5) ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€— βˆ’βˆ«1β–’γ€–[1∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€—]𝑑π‘₯+𝑐〗 yeβˆ’x = βˆ’(x βˆ’ 5)𝑒^(βˆ’π‘₯)βˆ’βˆ«1β–’γ€–γ€–βˆ’π‘’γ€—^(βˆ’π‘₯) 𝑑π‘₯γ€— + c yeβˆ’x = βˆ’(x βˆ’ 5)𝑒^(βˆ’π‘₯) + ∫1▒〖𝑒^(βˆ’π‘₯) 𝑑π‘₯γ€— + c Solution is y(IF) = ∫1β–’γ€–(𝑄×𝐼𝐹)𝑑π‘₯+𝑐〗 yex = ∫1β–’γ€–(5βˆ’π‘₯) 𝑒^π‘₯ 𝑑π‘₯+𝑐〗 yex = (5 – x) ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€— βˆ’ ∫1β–’[𝑑/𝑑π‘₯(5βˆ’π‘₯)∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€—]𝑑π‘₯ yex = (5 βˆ’ x) 𝑒^π‘₯βˆ’ ∫1β–’γ€–(βˆ’1)γ€— 𝑒^π‘₯ 𝑑π‘₯ yex = (5 βˆ’ x) 𝑒^π‘₯ + ∫1▒〖𝑒^π‘₯ 𝑑π‘₯γ€— Integrating by parts with ∫1β–’β–ˆ(𝑓(π‘₯) 𝑔(π‘₯) 𝑑π‘₯) =𝑓(π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯ γ€—βˆ’βˆ«1β–’γ€–[𝑓^β€² (π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯] 𝑑π‘₯γ€—γ€— Take f (x) = x – 5 & g (x) = 𝑒^(βˆ’π‘₯) Integrating by parts with ∫1β–’β–ˆ(𝑓(π‘₯) 𝑔(π‘₯) 𝑑π‘₯) =𝑓(π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯ γ€—βˆ’βˆ«1β–’γ€–[𝑓^β€² (π‘₯) ∫1▒〖𝑔(π‘₯) 𝑑π‘₯] 𝑑π‘₯γ€—γ€— Take f (x) = 5 – x & g (x) = 𝑒^π‘₯ ye–x =(5βˆ’π‘₯)𝑒^(βˆ’π‘₯) βˆ’ 𝑒^(βˆ’π‘₯)+𝑐 Dividing both sides by eβˆ’x y = (5 βˆ’x) βˆ’1 + cex y = 4 βˆ’ x + cex Since the curve passes through the point (0, 2) Put x = 0 & y = 2 2 = 4 βˆ’ 0 + ceΒ° 2 = 4 + C C = βˆ’2 ∴ Equation of curve is y = 4 βˆ’ x βˆ’ 2ex yex = (5 βˆ’ x) 𝑒^π‘₯+ 𝑒^π‘₯+𝑐 Dividing both sides by ex y = (5 βˆ’ x) + 1 + ceβˆ’x y = 6 βˆ’ x + ceβˆ’x Since curve passes through the point (0, 2) Put x = 0 & y = 2 2 = 6 βˆ’ 0 + ceΒ° 2 = 6 + C C = βˆ’4 ∴ Equation of curve is y = 6 βˆ’ x βˆ’ 4eβˆ’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.