# Example 23

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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

Example 1
Important

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7 Important

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13 Important

Example 14

Example 15

Example 16

Example 17 Important

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important

Example 23 You are here

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.