If y = ae 2x + be −x , then show that (d 2 y)/(dx 2 ) − dy/dx − 2y = 0

If y = ae^2x + be^−x , then show that y'' - y' - 2y = 0 - Teachoo

Question 22 - CBSE Class 12 Sample Paper for 2020 Boards - Part 2

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Transcript

Question 22 If y = ae2x + be−x , then show that (𝑑^2 𝑦)/(𝑑𝑥^2 ) − 𝑑𝑦/𝑑𝑥 − 2y = 0 Given 𝑦=𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥) Now, 𝑑𝑦/𝑑𝑥=2𝑎𝑒^2𝑥−𝑏𝑒^(−𝑥) And (𝑑^2 𝑦)/(𝑑𝑥^2 )=4𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥) Now, We need to show (𝑑^2 𝑦)/(𝑑𝑥^2 ) − 𝑑𝑦/𝑑𝑥 − 2y = 0 Solving LHS (𝑑^2 𝑦)/(𝑑𝑥^2 ) − 𝑑𝑦/𝑑𝑥 − 2y = (4𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥)) – (2𝑎𝑒^2𝑥−𝑏𝑒^(−𝑥)) – 2(𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥)) = 4𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥) – 2𝑎𝑒^2𝑥+𝑏𝑒^(−𝑥) – 2𝑎𝑒^2𝑥−2𝑏𝑒^(−𝑥) = (4𝑎𝑒^2𝑥 "– " 2𝑎𝑒^2𝑥 "– " 2𝑎𝑒^2𝑥)+(𝑏𝑒^(−𝑥) + 𝑏𝑒^(−𝑥) −2𝑏𝑒^(−𝑥)) = 0 + 0 = 0 Hence proved

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