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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.