Example 40 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Sept. 17, 2019 by Teachoo
Transcript
Example 40 If y = 3e2x + 2e3x, prove that ๐2๐ฆ๏ทฎ๐๐ฅ2๏ทฏ โ 5 ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ + 6y = 0. Given, ๐ฆ = 3๐2๐ฅ + 2๐3๐ฅ Differentiating ๐ค.๐.๐ก.๐ฅ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐ 3๐2๐ฅ + 2๐3๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = ๐ 3๐ 2๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ + ๐ 2๐ 3๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = 3. ๐2๐ฅ . ๐ 2๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ + 2 .๐ 3๐ฅ . ๐ 3๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = 3. ๐2๐ฅ . 2 + 2 .๐ 3๐ฅ. 3 ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = 6๐2๐ฅ + 6๐3๐ฅ ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = 6 (๐2๐ฅ + ๐3๐ฅ) Now, ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ = 6 (๐2๐ฅ + ๐3๐ฅ) Again Differentiating ๐ค.๐.๐ก.๐ฅ ๐๏ทฎ2๏ทฏ๐ฆ๏ทฎ ๐๐ฅ๏ทฎ2๏ทฏ๏ทฏ = ๐ 6 ๐2๐ฅ + ๐3๐ฅ๏ทฏ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ ๐๏ทฎ2๏ทฏ๐ฆ๏ทฎ ๐๐ฅ๏ทฎ2๏ทฏ๏ทฏ = 6 ๐ ๐2๐ฅ + ๐3๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ ๐๏ทฎ2๏ทฏ๐ฆ๏ทฎ ๐๐ฅ๏ทฎ2๏ทฏ๏ทฏ = 6 ๐ ๐2๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ + ๐ ๐3๐ฅ๏ทฏ๏ทฎ๐๐ฅ๏ทฏ๏ทฏ ๐๏ทฎ2๏ทฏ๐ฆ๏ทฎ ๐๐ฅ๏ทฎ2๏ทฏ๏ทฏ = 6 ๐2๐ฅ. 2+๐3๐ฅ.3๏ทฏ ๐ ๏ทฎ๐๏ทฏ๐๏ทฎ ๐ ๐๏ทฎ๐๏ทฏ๏ทฏ = 6 ๐๐๐๐+๐๐๐๐๏ทฏ Now we need to prove ๐2๐ฆ๏ทฎ๐๐ฅ2๏ทฏ โ 5 ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ + 6y = 0 Solving L.H.S ๐2๐ฆ๏ทฎ๐๐ฅ2๏ทฏ โ 5 ๐๐ฆ๏ทฎ๐๐ฅ๏ทฏ + 6y = 6 2๐2๐ฅ+3๐3๐ฅ๏ทฏ โ 5.6 ๐2๐ฅ+๐3๐ฅ๏ทฏ + 6 3๐2๐ฅ+2๐3๐ฅ๏ทฏ = 12๐2๐ฅ + 18๐3๐ฅ โ 30๐2๐ฅ โ 30๐3๐ฅ + 18๐2๐ฅ + 12๐3๐ฅ = 12๐2๐ฅ โ 30๐2๐ฅ + 18๐2๐ฅ + 18๐3๐ฅ โ 30๐3๐ฅ + 12๐3๐ฅ = 30๐2๐ฅ โ 30๐2๐ฅ + 30๐3๐ฅ โ 30๐3๐ฅ = 0 =RHS Hence proved
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.