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

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
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.