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Last updated at Dec. 24, 2018 by Teachoo
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Example 11 Find the particular solution of the differential equation ๐๐ฆ/๐๐ฅ=โ4๐ฅ๐ฆ^2 given that ๐ฆ=1 , ๐คโ๐๐ ๐ฅ=0 Given differential equation , ๐๐ฆ/๐๐ฅ=โ4๐ฅ๐ฆ^2 ๐๐ฆ/๐ฆ^2 = (โ4 x) dx Integrating both sides. โซ1โ๐๐ฆ/๐ฆ^2 = โซ1โใโ4๐ฅ ๐๐ฅใ โซ1โ๐๐ฆ/๐ฆ^2 = โ4 โซ1โใ๐ฅ ๐๐ฅใ ๐ฆ^(โ2+1)/(โ2+1) = โ4.๐ฅ^2/2 + c ๐ฆ^(โ1)/(โ1) = โ2x2 + c โ 1/๐ฆ = โ2x2 + c y = (โ1)/(โ2๐ฅ2 + ๐) y = (โ1)/(โ(2๐ฅ2 โ ๐)) y = 1/(2๐ฅ2 โ ๐) Given that at x = 0, y = 1 Putting x = 0, y = 1, in (1) 1 = 1/(2(0)^2 ) โ c 1 = 1/(โ๐ถ) c = โ1 Put c = โ1 in (1) y = 1/(2๐ฅ^2 ) โ(โ1) y = 1/(2๐ฅ^2 + 1) Hence, the particular solution of the equation is y = ๐/(๐๐^๐ + ๐)
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Example 28 Important Not in Syllabus - CBSE Exams 2021
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