Misc 2 (i) - Verify that the given function is a solution of the - Miscellaneous

part 2 - Misc 2 (i) - Miscellaneous - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Misc 2 (i) - Miscellaneous - Serial order wise - Chapter 9 Class 12 Differential Equations

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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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