Ex 5.7, 15 - If y = 500 e7x + 600 e-7x, show d2y/dx2 = 49y

Ex 5.7, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 15 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

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Transcript

Ex 5.7, 15 If 𝑦= 〖500𝑒〗^7𝑥+ 〖600𝑒〗^(−7𝑥), show that 𝑑2𝑦/𝑑𝑥2 = 49𝑦 𝑦= 〖500𝑒〗^7𝑥+ 〖600𝑒〗^(−7𝑥) Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = (𝑑(〖500𝑒〗^7𝑥 " +" 〖600𝑒〗^(−7𝑥) " " ))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = (𝑑 (〖500𝑒〗^7𝑥))/𝑑𝑥 + (𝑑 (〖600𝑒〗^(−7𝑥)))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 500 (𝑑 (𝑒^7𝑥))/𝑑𝑥 + 600 (𝑑 (𝑒^(−7𝑥)))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 500 . 𝑒^7𝑥. (𝑑 (7𝑥))/𝑑𝑥 + 600 . 𝑒^(−7𝑥) . (𝑑 (−7𝑥))/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 500 . 𝑒^7𝑥. 7 + 600 . 𝑒^(−7𝑥) . (−7) 𝑑𝑦/𝑑𝑥 = 500 . 7 . 𝑒^7𝑥 − 600 . 7 . 𝑒^(−7𝑥) Again Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑/𝑑𝑥 (𝑑𝑦/𝑑𝑥) = 𝑑(500 . 7 . 𝑒^7𝑥 " −" 〖 600 . 7 . 𝑒〗^(−7𝑥) )" " /𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 𝑑(500 × 7 . 𝑒^7𝑥 )/𝑑𝑥 − 𝑑(〖600 × 7𝑒〗^(−7𝑥) )" " /𝑑𝑥 (𝑑^2 𝑦)/(𝑑𝑥^2 ) = 500 × 7 𝑑(𝑒^7𝑥 )/𝑑𝑥 − 600 × 7 𝑑(𝑒^(−7𝑥) )/𝑑𝑥 = 500 × 7 × 𝑒^7𝑥 𝑑(7𝑥)/𝑑𝑥 − 600 × 7 × 𝑒^(−7𝑥) 𝑑(−7𝑥)/𝑑𝑥 = 500 × 7 × 𝑒^7𝑥. 7 − 600 × 7 × 𝑒^(−7𝑥) (−7) = 500 × 7 × 7𝑒^7𝑥 + 600 × 7 × 7 × 𝑒^(−7𝑥) = 7 × 7 (500〖 𝑒〗^7𝑥+600〖 𝑒〗^(−7𝑥) ) = 49 (500〖 𝑒〗^7𝑥+600〖 𝑒〗^(−7𝑥) ) = 49 𝑦 Hence proved . (As 𝑦= 〖500𝑒〗^7𝑥+ 〖600𝑒〗^(−7𝑥))

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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.