Example 7 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Example 7 Find the equation of the curve passing through the point (1 , 1) whose differential equation is ๐ฅ ๐๐ฆ= (2๐ฅ^2+1)๐๐ฅ(๐ฅโ 0) ๐ฅ ๐๐ฆ = (2x2 + 1)dx dy = "(2x2 + 1)" /๐ฅ dx dy = ("2x2" /๐ฅ+1/๐ฅ) dx dy = (๐๐+๐/๐) dx Integrating both sides. โซ1โ๐๐ฆ = โซ1โ(2๐ฅ+1/๐ฅ) ๐๐ฅ โซ1โ๐๐ฆ = โซ1โใ2๐ฅ ๐๐ฅ+ใ โซ1โใ1/๐ฅ ๐๐ฅใ y = 2 ๐ฅ2/2 + log |๐ฅ| + C y = ๐๐ + log |๐| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |๐| + C 1 = 1 + 0 + C 1 โ 1 = C โด C = 0 Put C = 0 in (1) y = x2 + log |๐ฅ| + C y = x2 + log |๐| + 0 y = x2 + log |๐ฅ| Hence, the equation of curve is y = x2 + log |๐|
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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