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Example 8 Find the equation of a curve passing through the point (โˆ’2 ,3), given that the slope of the tangent to the curve at any point (๐‘ฅ , ๐‘ฆ) is 2๐‘ฅ/๐‘ฆ^2 Slope of tangent = ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ โˆด ๐’…๐’š/๐’…๐’™ = ๐Ÿ๐’™/๐’š๐Ÿ ๐‘ฆ2 dy = 2x dx Integrating both sides โˆซ1โ–’๐’š๐Ÿ ๐’…๐’š= โˆซ1โ–’ใ€–๐Ÿ๐’™ ๐’…๐’™ใ€— ๐‘ฆ^3/3 = 2.๐‘ฅ^2/2 + C ๐‘ฆ^3/3 = ๐‘ฅ^2 + C ๐‘ฆ^3 = ใ€–3๐‘ฅใ€—^2+3๐ถ ๐’š^๐Ÿ‘ = ใ€–๐Ÿ‘๐’™ใ€—^๐Ÿ+๐‘ช๐Ÿ where ๐ถ1 = 3C Given that equation passes through (โˆ’2, 3) Putting x = โˆ’2, y = 3 in (1) y3 = 3x2 + C1 33 = 3(โˆ’2)2 + C1 27 = 3 ร— 4 + C1 27 โˆ’ 12 = C1 15 = C1 C1 = 15 Putting C1 in (1) y3 = 3x2 + 15 y = "(3x2 + " ใ€–"15)" ใ€—^(๐Ÿ/๐Ÿ‘) " "is the particular solution of the equation.

  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo