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Ex 5.7, 14 - If y = A emx + B enx, show d2y/dx2 - (m + n)

Ex 5.7, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 2
Ex 5.7, 14 - Chapter 5 Class 12 Continuity and Differentiability - Part 3

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Ex 5.7, 14 If 𝑦= γ€–A𝑒〗^π‘šπ‘₯ + γ€–B𝑒〗^𝑛π‘₯, show that 𝑑2𝑦/𝑑π‘₯2 βˆ’ (π‘š+𝑛) 𝑑𝑦/𝑑π‘₯ + π‘šπ‘›π‘¦ = 0 𝑦= γ€–A𝑒〗^π‘šπ‘₯ + γ€–B𝑒〗^𝑛π‘₯ Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(γ€–A𝑒〗^π‘šπ‘₯ " + " γ€–B𝑒〗^𝑛π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑑(γ€–A𝑒〗^π‘šπ‘₯))/𝑑π‘₯ + (𝑑(γ€–B𝑒〗^𝑛π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = A . 𝑒^π‘šπ‘₯. (𝑑(π‘šπ‘₯))/𝑑π‘₯ + B . 𝑒^𝑛π‘₯ (𝑑(𝑛π‘₯))/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = A . 𝑒^π‘šπ‘₯. π‘š + B . 𝑒^𝑛π‘₯. 𝑛 𝑑𝑦/𝑑π‘₯ = π΄π‘šπ‘’^π‘šπ‘₯ + 𝐡𝑛𝑒^𝑛π‘₯ Again Differentiating 𝑀.π‘Ÿ.𝑑.π‘₯ 𝑑/𝑑π‘₯ (𝑑𝑦/𝑑π‘₯) = 𝑑(π΄π‘šπ‘’^π‘šπ‘₯ " + " 𝐡𝑛𝑒^𝑛π‘₯ " " )" " /𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = 𝑑(π΄π‘šπ‘’^π‘šπ‘₯ )/𝑑π‘₯ + 𝑑(𝐡𝑛𝑒^𝑛π‘₯ )" " /𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = π΄π‘š 𝑑(𝑒^π‘šπ‘₯ )/𝑑π‘₯ + 𝐡𝑛 𝑑(𝑒^𝑛π‘₯ )" " /𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = π΄π‘š . 𝑒^(π‘šπ‘₯ ). 𝑑(π‘šπ‘₯ )/𝑑π‘₯ + 𝐡𝑛 . 𝑒^𝑛π‘₯ . 𝑑(𝑛π‘₯)/𝑑π‘₯ (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = π΄π‘šπ‘’^(π‘šπ‘₯ ) . π‘š+𝐡𝑛𝑒^𝑛π‘₯ . 𝑛 (𝑑^2 𝑦)/(𝑑π‘₯^2 ) = π΄π‘š2𝑒^(π‘šπ‘₯ )+𝐡𝑛2𝑒^𝑛π‘₯ We need to prove (𝑑^2 𝑦)/(𝑑π‘₯^2 ) βˆ’ (π‘š+𝑛) 𝑑𝑦/𝑑π‘₯ + π‘šπ‘›π‘¦ = 0 Solving LHS (𝑑^2 𝑦)/(𝑑π‘₯^2 ) βˆ’ (π‘š+𝑛) 𝑑𝑦/𝑑π‘₯ + π‘šπ‘›π‘¦ = (π΄π‘š2𝑒^(π‘šπ‘₯ )+𝐡𝑛2𝑒^𝑛π‘₯) βˆ’ (π‘š+𝑛) (π΄π‘šπ‘’^(π‘šπ‘₯ )+𝐡𝑛𝑒^𝑛π‘₯) + π‘šπ‘› (𝐴𝑒^(π‘šπ‘₯ )+𝐡𝑒^𝑛π‘₯) = π΄π‘š2𝑒^(π‘šπ‘₯ )+𝐡𝑛2𝑒^𝑛π‘₯ βˆ’ π‘š(π΄π‘šπ‘’^(π‘šπ‘₯ )+𝐡𝑛𝑒^𝑛π‘₯) βˆ’ 𝑛(π΄π‘šπ‘’^(π‘šπ‘₯ )+𝐡𝑛𝑒^𝑛π‘₯) + π‘šπ‘› 𝐴𝑒^(π‘šπ‘₯ )+π‘šπ‘›π΅π‘’^𝑛π‘₯ = π΄π‘š2𝑒^(π‘šπ‘₯ )+𝐡𝑛2𝑒^𝑛π‘₯ βˆ’π΄π‘š2𝑒^(π‘šπ‘₯ )βˆ’ π΅π‘šπ‘›π‘’^𝑛π‘₯ βˆ’ π΄π‘›π‘šπ‘’^(π‘šπ‘₯ ) + 𝐡𝑛2𝑒^𝑛π‘₯+ π‘šπ‘› 𝐴𝑒^(π‘šπ‘₯ )+π‘šπ‘›π΅π‘’^𝑛π‘₯ = π΄π‘š2𝑒^(π‘šπ‘₯ )βˆ’ π΄π‘š2𝑒^(π‘šπ‘₯ ) + 𝐡𝑛2𝑒^𝑛π‘₯ βˆ’π΅π‘›2𝑒^𝑛π‘₯ βˆ’ π΅π‘šπ‘›π‘’^𝑛π‘₯ + π΅π‘šπ‘›π‘’^𝑛π‘₯ βˆ’ π΄π‘›π‘šπ‘’^(π‘šπ‘₯ ) + π΄π‘›π‘šπ‘’^(π‘šπ‘₯ ) = 0 = RHS Hence proved

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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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.