# Ex 9.4, 17 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 10, 2019 by Teachoo

Last updated at Dec. 10, 2019 by Teachoo

Transcript

Ex 9.4, 17 Find the equation of a curve passing through the point (0 , โ2) , given that at any point (๐ฅ , ๐ฆ) on the curve , the product of the slope of its tangent and ๐ฆ coordinate of the point is equal to the ๐ฅ coordinate of the point . Slope of tangent to the curve = ๐๐ฆ/๐๐ฅ 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 ๐๐ฆ/๐๐ฅ = x y dy = x dx โฆ(1) Integrating both sides โซ1โใ๐ฆ ๐๐ฆ=โซ1โใ๐ฅ ๐๐ฅ ใ ใ ๐ฆ^2/2 = ๐ฅ^2/2 + C The curve passes through point (0, โ2) Putting x = 0 & y = โ2 in equation (โ2)^2/2 = 0^2/2 + C 4/2 = C C = 2 Putting back value of C in equation ๐ฆ^2/2 = ๐ฅ^2/2 + 2 ๐ฆ^2/2 = (๐ฅ^2 + 4)/2 ๐ฆ^2 = ๐ฅ^2 + 4 ๐ฆ^2 โ ๐ฅ^2= 4 Hence, equation of the curve is ๐^๐ โ ๐^๐= 4

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Chapter 9 Class 12 Differential Equations (Term 2)

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.