# Example 23 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 23 Find the equation of a curve passing the point 0 , 1. if the slope of the tangent to the curve at any point 𝑥, 𝑦 is equal to the sum of the 𝑥 coordinate 𝑎𝑏𝑠𝑐𝑖𝑠𝑠𝑎 and the product of the coordinate and 𝑦 coordinate 𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 of that point . Slope of the tangent to the curve at (x, y) = 𝑑𝑦𝑑𝑥 Given equation: 𝑑𝑦𝑑𝑥 = x + xy 𝑑𝑦𝑑𝑥 − xy = x Differential equation is of the form 𝑑𝑦𝑑𝑥 + Py = Q where P = –x & Q = x IF = e 𝑝𝑑𝑥 = e− 𝑥 𝑑𝑥 = e −𝑥22 Solution is y (IF) = 𝑄×𝐼𝐹𝑑𝑥+𝐶 y 𝑒 − 𝑥22 = 𝑥 𝑒 −𝑥22 𝑑𝑥+𝐶 Putting − 𝑥22 = t − 2𝑥2 𝑑𝑥=𝑑𝑡 x dx = − dt Thus, our equations become y e − 𝑥22 = − 𝑒𝑡𝑑𝑡+𝑐 y e − 𝑥22 = − 𝑒𝑡+𝑐 Putting t = − 𝑥22 y e − 𝑥22 = − e − 𝑥22 + C 𝑦 𝑒 𝑥22 = − 1 𝑒 𝑥22 + C y = −1 + C e 𝑥22 Since curve passes through (0, 1) Putting x = 0, y = 1 in (1) y = −1 + C 𝑒 𝑥22 1 = −1 + C e 022 1 = −1 + C. 1 + 1 = C C = 2 Putting value of C in (1) y = −1 + C e 𝑥22 y = −1 + 2 e 𝑥22 ∴ Equation of the curve is y = −1 + 𝟐 𝒆 𝒙𝟐𝟐

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Chapter 9 Class 12 Differential Equations

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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.