Ex 5.7, 15 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)

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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))