# Example 13

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Example 13Find the equation of a curve passing through the point (โ2 ,3), given that the slope of the tangent to the curve at any point (๐ฅ , ๐ฆ) is 2๐ฅ/๐ฆ^2 Slope of tangent = ๐๐ฆ/๐๐ฅ โด ๐๐ฆ/๐๐ฅ = 2๐ฅ/๐ฆ2 ๐ฆ2 dy = 2x dx Integrating both sides โซ1โ๐ฆ2 ๐๐ฆ= โซ1โใ2๐ฅ ๐๐ฅใ ๐ฆ^3/3 = 2.๐ฅ^2/2 + C ๐ฆ^3/3 = ๐ฅ^2 + C ๐ฆ^3 = ใ3๐ฅใ^2+3๐ถ ๐ฆ^3 = ใ3๐ฅใ^2+๐ถ1 where ๐ถ1 = 3C Given that equation passes through (โ2, 3) Putting x = โ2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(โ2)2 + C1 27 = 3 ร 4 + C1 27 โ 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " ใ"15)" ใ^(๐/๐) " "is the particular solution of the equation.

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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.