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Gen and Particular Solution
Gen and Particular Solution
Last updated at August 13, 2026 by Teachoo
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Transcript
Misc 2 For each of the exercise given below , verify that the given function (๐๐๐๐๐๐๐๐ก ๐๐ ๐๐ฅ๐๐๐๐๐๐ก) is a solution of the corresponding differential equation . (i) ๐ฅ๐ฆ=๐ ๐^๐ฅ+๐ ๐^(โ๐ฅ)+๐ฅ^2 : ๐ฅ (๐^2 ๐ฆ)/(๐๐ฅ^2 )+2 ๐๐ฆ/๐๐ฅโ๐ฅ๐ฆ+๐ฅ^2โ2=0 ๐ฅ๐ฆ=๐๐^๐ฅ+๐ ๐^(โ๐ฅ)+๐ฅ^2 Differentiating w.r.t x (๐(๐ฅ๐ฆ))/๐๐ฅ=๐/๐๐ฅ [๐ ๐^๐ฅ+๐ ๐^(โ๐ฅ)+๐ฅ^2 ] ๐๐ฅ/๐๐ฅ y+๐๐ฆ/๐๐ฅ ๐ฅ =๐ใ ๐ใ^๐ฅ+(โ1)๐ ๐^(โ๐ฅ)+2๐ฅ ๐+๐^โฒ ๐ =๐ใ ๐ใ^๐โ๐ ๐^(โ๐)+๐๐ Differentiating again w.r.t x ๐ฆโฒ+(๐ฆ^โฒ ๐ฅ)^โฒ =(๐๐^๐ฅ )^โฒโ(๐๐^(โ๐ฅ) )^โฒ+(2๐ฅ)^โฒ ๐ฆ^โฒ+(๐ฆ^โฒโฒ ๐ฅ+๐ฆ^โฒร1)=๐๐^๐ฅ+๐๐^(โ๐ฅ)+2 ๐^โฒโฒ ๐+๐๐^โฒ=๐๐^๐+๐๐^(โ๐)+๐ Now, we know that ๐ฅ๐ฆ=๐๐^๐ฅ+๐ ๐^(โ๐ฅ)+๐ฅ^2 ๐ฅ๐ฆโ๐ฅ^2=๐๐^๐ฅ+๐ ๐^(โ๐ฅ) ๐๐^๐+๐ ๐^(โ๐)=๐๐โ๐^๐ Putting (2) in (1) ๐ฆ^โฒโฒ ๐ฅ+2๐ฆ^โฒ=๐๐^๐+๐๐^(โ๐)+2 ๐ฆ^โฒโฒ ๐ฅ+2๐ฆ^โฒ=๐๐โ๐^๐+2 (๐^2 ๐ฆ)/(๐๐ฅ^2 ) ๐ฅ+2 ๐๐ฆ/๐๐ฅ=๐๐โ๐^๐+2 (๐ ^๐ ๐)/(๐ ๐^๐ ) ๐+๐ ๐ ๐/๐ ๐โ๐๐+๐^๐=๐ โด The given function is a solution