# Example 12 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 12Find 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 = (2๐ฅ+1/๐ฅ) 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 = ๐ฅ2 + log |๐ฅ| + 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) i.e y = x2 + log |๐ฅ| + C y = x2 + log |๐ฅ| + 0 y = x2 + log |๐ฅ| Hence, the equation of curve is y = x2 + log |๐|

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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.