# Ex 9.6, 16

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

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 𝑥 , 𝑦 is equal to the sum of the coordinates of the point. We know that slope of tangent to the curve = 𝑑𝑦𝑑𝑥 at (x, y) Given 𝑑𝑦𝑑𝑥 = x + y 𝑑𝑦𝑑𝑥 − y = x This is the form 𝑑𝑦𝑑𝑥+Py=Q where P = −1 & Q = x Finding Integrating factor IF = 𝑒 𝑝𝑑𝑥 IF = 𝑒 (−1)𝑑𝑥 IF = e−x Solution is y(IF) = 𝑄×𝐼𝐹𝑑𝑥+𝑐 𝑦 𝑒−𝑥 = 𝑥 𝑒−𝑥 𝑑𝑥+𝑐 ye−x = x 𝑒−𝑥 𝑑𝑥− 1 𝑒−𝑥 𝑑𝑥 𝑑𝑥+𝑐 ye−x = −x 𝑒−𝑥 − − 𝑒−𝑥𝑑𝑥+𝑐 ye−x = −x 𝑒−𝑥+ 𝑒−𝑥𝑑𝑥+𝑐 ye−x = −x 𝑒−𝑥+ − 𝑒−𝑥−1 +𝑐 ye−x = −x 𝑒−𝑥− 𝑒−𝑥 +𝑐 Dividing both sides by e−x y = −x − 1 + 𝑐 𝑒−𝑥 y = −x − 1 + cex Since curve passes through origin, Putting x = 0 & y = 0 in (2) 0 = 0 − 1 + Ce0 1 = C C = 1 Put value of C in (1) y = −x − 1 + ex x + y + 1 = ex

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.