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Transcript

Misc 2 For each of the exercise given below , verify that the given function (๐‘–๐‘š๐‘๐‘™๐‘–๐‘๐‘–๐‘ก ๐‘œ๐‘Ÿ ๐‘’๐‘ฅ๐‘๐‘™๐‘–๐‘๐‘–๐‘ก) is a solution of the corresponding differential equation . (i) ๐‘ฅ๐‘ฆ=๐‘Ž ๐‘’^๐‘ฅ+๐‘ ๐‘’^(โˆ’๐‘ฅ)+๐‘ฅ^2 : ๐‘ฅ (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 )+2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅโˆ’๐‘ฅ๐‘ฆ+๐‘ฅ^2โˆ’2=0 ๐‘ฅ๐‘ฆ=๐‘Ž๐‘’^๐‘ฅ+๐‘ ๐‘’^(โˆ’๐‘ฅ)+๐‘ฅ^2 Differentiating w.r.t x (๐‘‘(๐‘ฅ๐‘ฆ))/๐‘‘๐‘ฅ=๐‘‘/๐‘‘๐‘ฅ [๐‘Ž ๐‘’^๐‘ฅ+๐‘ ๐‘’^(โˆ’๐‘ฅ)+๐‘ฅ^2 ] ๐‘‘๐‘ฅ/๐‘‘๐‘ฅ y+๐‘‘๐‘ฆ/๐‘‘๐‘ฅ ๐‘ฅ =๐‘Žใ€– ๐‘’ใ€—^๐‘ฅ+(โˆ’1)๐‘ ๐‘’^(โˆ’๐‘ฅ)+2๐‘ฅ ๐’š+๐’š^โ€ฒ ๐’™ =๐’‚ใ€– ๐’†ใ€—^๐’™โˆ’๐’ƒ ๐’†^(โˆ’๐’™)+๐Ÿ๐’™ Differentiating again w.r.t x ๐‘ฆโ€ฒ+(๐‘ฆ^โ€ฒ ๐‘ฅ)^โ€ฒ =(๐‘Ž๐‘’^๐‘ฅ )^โ€ฒโˆ’(๐‘๐‘’^(โˆ’๐‘ฅ) )^โ€ฒ+(2๐‘ฅ)^โ€ฒ ๐‘ฆ^โ€ฒ+(๐‘ฆ^โ€ฒโ€ฒ ๐‘ฅ+๐‘ฆ^โ€ฒร—1)=๐‘Ž๐‘’^๐‘ฅ+๐‘๐‘’^(โˆ’๐‘ฅ)+2 ๐’š^โ€ฒโ€ฒ ๐’™+๐Ÿ๐’š^โ€ฒ=๐’‚๐’†^๐’™+๐’ƒ๐’†^(โˆ’๐’™)+๐Ÿ Now, we know that ๐‘ฅ๐‘ฆ=๐‘Ž๐‘’^๐‘ฅ+๐‘ ๐‘’^(โˆ’๐‘ฅ)+๐‘ฅ^2 ๐‘ฅ๐‘ฆโˆ’๐‘ฅ^2=๐‘Ž๐‘’^๐‘ฅ+๐‘ ๐‘’^(โˆ’๐‘ฅ) ๐’‚๐’†^๐’™+๐’ƒ ๐’†^(โˆ’๐’™)=๐’™๐’šโˆ’๐’™^๐Ÿ Putting (2) in (1) ๐‘ฆ^โ€ฒโ€ฒ ๐‘ฅ+2๐‘ฆ^โ€ฒ=๐’‚๐’†^๐’™+๐’ƒ๐’†^(โˆ’๐’™)+2 ๐‘ฆ^โ€ฒโ€ฒ ๐‘ฅ+2๐‘ฆ^โ€ฒ=๐’™๐’šโˆ’๐’™^๐Ÿ+2 (๐‘‘^2 ๐‘ฆ)/(๐‘‘๐‘ฅ^2 ) ๐‘ฅ+2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐’™๐’šโˆ’๐’™^๐Ÿ+2 (๐’…^๐Ÿ ๐’š)/(๐’…๐’™^๐Ÿ ) ๐’™+๐Ÿ ๐’…๐’š/๐’…๐’™โˆ’๐’™๐’š+๐’™^๐Ÿ=๐ŸŽ โˆด The given function is a solution

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.