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  1. Chapter 9 Class 12 Differential Equations
  2. Concept wise

Transcript

Ex 9.4, 17 Find the equation of a curve passing through the point (0 , โˆ’2) , given that at any point (๐‘ฅ , ๐‘ฆ) on the curve , the product of the slope of its tangent and ๐‘ฆ coordinate of the point is equal to the ๐‘ฅ coordinate of the point . Slope of tangent to the curve = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ Given at any point (x, y), product of slope of its tangent and y-coordinate is equal to x-coordinate of the point Therefore, y ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = x y dy = x dx โ€ฆ(1) Integrating both sides โˆซ1โ–’ใ€–๐‘ฆ ๐‘‘๐‘ฆ=โˆซ1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅ ใ€— ใ€— ๐‘ฆ^2/2 = ๐‘ฅ^2/2 + C The curve passes through point (0, โˆ’2) Putting x = 0 & y = โˆ’2 in equation (โˆ’2)^2/2 = 0^2/2 + C 4/2 = C C = 2 Putting back value of C in equation ๐‘ฆ^2/2 = ๐‘ฅ^2/2 + 2 ๐‘ฆ^2/2 = (๐‘ฅ^2 + 4)/2 ๐‘ฆ^2 = ๐‘ฅ^2 + 4 ๐‘ฆ^2 โˆ’ ๐‘ฅ^2= 4 Hence, equation of the curve is ๐’š^๐Ÿ โˆ’ ๐’™^๐Ÿ= 4

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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.