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Ex 5.7, 14 - If y = A emx + B enx, show d2y/dx2 - (m + n) - Finding second order derivatives- Implicit form

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  1. Chapter 5 Class 12 Continuity and Differentiability
  2. Serial order wise
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Ex 5.7, 14 If ๐‘ฆ= A๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ + B๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ, show that ๐‘‘2๐‘ฆ๏ทฎ๐‘‘๐‘ฅ2๏ทฏ โˆ’ ๐‘š+๐‘›๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐‘š๐‘›๐‘ฆ = 0 ๐‘ฆ= A๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ + B๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ = ๐‘‘( A๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ + B๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ)๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ = ๐‘‘( A๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ)๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐‘‘( B๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ)๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ = A . ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ. ๐‘‘(๐‘š๐‘ฅ)๏ทฎ๐‘‘๐‘ฅ๏ทฏ + B . ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ ๐‘‘(๐‘›๐‘ฅ)๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ = A . ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ. ๐‘š + B . ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ. ๐‘› ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ = ๐ด๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ + ๐ต๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ Again Differentiating ๐‘ค.๐‘Ÿ.๐‘ก.๐‘ฅ ๐‘‘๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ๏ทฏ = ๐‘‘ ๐ด๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ + ๐ต๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ ๏ทฏ ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐‘‘ ๐ด๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ ๏ทฏ๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐‘‘ ๐ต๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ๏ทฏ ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐ด๐‘š ๐‘‘ ๐‘’๏ทฎ๐‘š๐‘ฅ๏ทฏ ๏ทฏ๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐ต๐‘› ๐‘‘ ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ๏ทฏ ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐ด๐‘š . ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ. ๐‘‘ ๐‘š๐‘ฅ ๏ทฏ๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐ต๐‘› . ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ . ๐‘‘ ๐‘›๐‘ฅ๏ทฏ๏ทฎ๐‘‘๐‘ฅ๏ทฏ ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐ด๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ . ๐‘š+๐ต๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ . ๐‘› ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐ด๐‘š2 ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ+๐ต๐‘›2 ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ Hence ๐’…๏ทฎ๐Ÿ๏ทฏ๐’š๏ทฎ๐’… ๐’™๏ทฎ๐Ÿ๏ทฏ๏ทฏ = ๐‘จ๐’Ž๐Ÿ ๐’†๏ทฎ๐’Ž๐’™ ๏ทฏ+๐‘ฉ๐’๐Ÿ ๐’†๏ทฎ๐’๐’™๏ทฏ We need to prove ๐‘‘๏ทฎ2๏ทฏ๐‘ฆ๏ทฎ๐‘‘ ๐‘ฅ๏ทฎ2๏ทฏ๏ทฏ โˆ’ (๐‘š+๐‘›) ๐‘‘๐‘ฆ๏ทฎ๐‘‘๐‘ฅ๏ทฏ + ๐‘š๐‘›๐‘ฆ = 0 Solving LHS = (๐ด๐‘š2 ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ+๐ต๐‘›2 ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ) โˆ’ (๐‘š+๐‘›) (๐ด๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ+๐ต๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ) + ๐‘š๐‘› (๐ด ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ+๐ต ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ) = Am2 e๏ทฎmx ๏ทฏ+Bn2 e๏ทฎnx๏ทฏ โˆ’ ๐‘š(Am e๏ทฎmx ๏ทฏ+Bn e๏ทฎnx๏ทฏ) โˆ’ ๐‘›(Am e๏ทฎmx ๏ทฏ+Bn e๏ทฎnx๏ทฏ) + ๐‘š๐‘› A e๏ทฎmx ๏ทฏ+๐‘š๐‘›B e๏ทฎnx๏ทฏ = Am2 e๏ทฎmx ๏ทฏ+Bn2 e๏ทฎnx๏ทฏ โˆ’Am2 e๏ทฎmx ๏ทฏโˆ’ Bmn e๏ทฎnx๏ทฏ โˆ’ Anm e๏ทฎmx ๏ทฏ + Bn2 e๏ทฎnx๏ทฏ+ ๐‘š๐‘› A e๏ทฎmx ๏ทฏ+๐‘š๐‘›B e๏ทฎnx๏ทฏ = ๐ด๐‘š2 ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏโˆ’ ๐ด๐‘š2 ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ + ๐ต๐‘›2 ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ โˆ’๐ต๐‘›2 ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ โˆ’ ๐ต๐‘š๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ + ๐ต๐‘š๐‘› ๐‘’๏ทฎ๐‘›๐‘ฅ๏ทฏ โˆ’ ๐ด๐‘›๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ + ๐ด๐‘›๐‘š ๐‘’๏ทฎ๐‘š๐‘ฅ ๏ทฏ = 0 + 0 + 0 + 0 + 0 = 0 = RHS Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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