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

Transcript

Example 11 Find the particular solution of the differential equation ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’4๐‘ฅ๐‘ฆ^2 given that ๐‘ฆ=1 , ๐‘คโ„Ž๐‘’๐‘› ๐‘ฅ=0 Given differential equation , ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=โˆ’4๐‘ฅ๐‘ฆ^2 ๐‘‘๐‘ฆ/๐‘ฆ^2 = (โˆ’4 x) dx Integrating both sides. โˆซ1โ–’๐‘‘๐‘ฆ/๐‘ฆ^2 = โˆซ1โ–’ใ€–โˆ’4๐‘ฅ ๐‘‘๐‘ฅใ€— โˆซ1โ–’๐‘‘๐‘ฆ/๐‘ฆ^2 = โˆ’4 โˆซ1โ–’ใ€–๐‘ฅ ๐‘‘๐‘ฅใ€— ๐‘ฆ^(โˆ’2+1)/(โˆ’2+1) = โˆ’4.๐‘ฅ^2/2 + c ๐‘ฆ^(โˆ’1)/(โˆ’1) = โˆ’2x2 + c โˆ’ 1/๐‘ฆ = โ€“2x2 + c y = (โˆ’1)/(โˆ’2๐‘ฅ2 + ๐‘) y = (โˆ’1)/(โˆ’(2๐‘ฅ2 โˆ’ ๐‘)) y = 1/(2๐‘ฅ2 โˆ’ ๐‘) Given that at x = 0, y = 1 Putting x = 0, y = 1, in (1) 1 = 1/(2(0)^2 ) โˆ’ c 1 = 1/(โˆ’๐ถ) c = โˆ’1 Put c = โˆ’1 in (1) y = 1/(2๐‘ฅ^2 ) โˆ’(โˆ’1) y = 1/(2๐‘ฅ^2 + 1) Hence, the particular solution of the equation is y = ๐Ÿ/(๐Ÿ๐’™^๐Ÿ + ๐Ÿ)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.