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Misc 2 - Verify given function is a solution of differential

Misc 2 - Chapter 9 Class 12 Differential Equations - Part 2
Misc 2 - Chapter 9 Class 12 Differential Equations - Part 3

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Misc 2 For each of the exercise given below , verify that the given function (š‘–š‘šš‘š‘™š‘–š‘š‘–š‘” š‘œš‘Ÿ š‘’š‘„š‘š‘™š‘–š‘š‘–š‘”) is a solution of the corresponding differential equation . (i) š‘„š‘¦=š‘Ž š‘’^š‘„+š‘ š‘’^(āˆ’š‘„)+š‘„^2 : š‘„ (š‘‘^2 š‘¦)/(š‘‘š‘„^2 )+2 š‘‘š‘¦/š‘‘š‘„āˆ’š‘„š‘¦+š‘„^2āˆ’2=0 š‘„š‘¦=š‘Žš‘’^š‘„+š‘ š‘’^(āˆ’š‘„)+š‘„^2 Differentiating w.r.t x (š‘‘(š‘„š‘¦))/š‘‘š‘„=š‘‘/š‘‘š‘„ [š‘Ž š‘’^š‘„+š‘ š‘’^(āˆ’š‘„)+š‘„^2 ] š‘‘š‘„/š‘‘š‘„ y+š‘‘š‘¦/š‘‘š‘„ š‘„ =š‘Žć€– š‘’怗^š‘„+(āˆ’1)š‘ š‘’^(āˆ’š‘„)+2š‘„ y+y^ā€² x =a怖 e怗^xāˆ’b e^(āˆ’x)+2x Differentiating again w.r.t x š‘¦ā€²+(š‘¦^ā€² š‘„)^ā€² =(š‘Žš‘’^š‘„ )^ā€²āˆ’(š‘š‘’^(āˆ’š‘„) )^ā€²+(2š‘„)^ā€² š‘¦^ā€²+(š‘¦^ā€²ā€² š‘„+š‘¦^ā€²Ć—1)=š‘Žš‘’^š‘„+š‘š‘’^(āˆ’š‘„)+2 š‘¦^ā€²ā€² š‘„+2š‘¦^ā€²=š‘Žš‘’^š‘„+š‘š‘’^(āˆ’š‘„)+2 Now, we know that š‘„š‘¦=š‘Žš‘’^š‘„+š‘ š‘’^(āˆ’š‘„)+š‘„^2 š‘„š‘¦āˆ’š‘„^2=š‘Žš‘’^š‘„+š‘ š‘’^(āˆ’š‘„) š‘Žš‘’^š‘„+š‘ š‘’^(āˆ’š‘„)=š‘„š‘¦āˆ’š‘„^2 ...(1) (Given equation) ...(2) Putting (2) in (1) š‘¦^ā€²ā€² š‘„+2š‘¦^ā€²=š’‚š’†^š’™+š’ƒš’†^(āˆ’š’™)+2 š‘¦^ā€²ā€² š‘„+2š‘¦^ā€²=š’™š’šāˆ’š’™^šŸ+2 (š‘‘^2 š‘¦)/(š‘‘š‘„^2 ) š‘„+2 š‘‘š‘¦/š‘‘š‘„=š’™š’šāˆ’š’™^šŸ+2 (š’…^šŸ š’š)/(š’…š’™^šŸ ) š’™+šŸ š’…š’š/š’…š’™āˆ’š’™š’š+š’™^šŸ=šŸŽ āˆ“ The given function is a solution

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.