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Example 12 - Find equation: (1, 1) , x dy = (2x2 + 1) dx - Examples

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Example 12 Find the equation of the curve passing through the point (1 , 1) whose differential equation is ๐‘ฅ ๐‘‘๐‘ฆ= (2๐‘ฅ^2+1)๐‘‘๐‘ฅ(๐‘ฅโ‰ 0) ๐‘ฅ ๐‘‘๐‘ฆ = (2x2 + 1)dx dy = "(2x2 + 1)" /๐‘ฅ dx dy = ("2x2" /๐‘ฅ+1/๐‘ฅ) dx dy = (2๐‘ฅ+1/๐‘ฅ) dx Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’(2๐‘ฅ+1/๐‘ฅ) ๐‘‘๐‘ฅ โˆซ1โ–’๐‘‘๐‘ฆ = โˆซ1โ–’ใ€–2๐‘ฅ ๐‘‘๐‘ฅ+ใ€— โˆซ1โ–’ใ€–1/๐‘ฅ ๐‘‘๐‘ฅใ€— y = 2 ๐‘ฅ2/2 + log |๐‘ฅ| + C y = ๐‘ฅ2 + log |๐‘ฅ| + C Since the curve passes through point (1, 1) Putting x = 1, y = 1 is (1) 1 = 12 + log |๐Ÿ| + C 1 = 1 + 0 + C 1 โˆ’ 1 = C โˆด C = 0 Put C = 0 in (1) i.e y = x2 + log |๐‘ฅ| + C y = x2 + log |๐‘ฅ| + 0 y = x2 + log |๐‘ฅ| Hence, the equation of curve is y = x2 + log |๐’™|

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