Example 40 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at April 13, 2021 by Teachoo
Transcript
Example 40 If y = 3e2x + 2e3x, prove that ๐2๐ฆ/๐๐ฅ2 โ 5 ๐๐ฆ/๐๐ฅ + 6y = 0. Given, ๐ฆ = 3๐2๐ฅ + 2๐3๐ฅ Differentiating ๐ค.๐.๐ก.๐ฅ ๐๐ฆ/๐๐ฅ = ๐(3๐2๐ฅ + 2๐3๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ = ๐(3๐ 2๐ฅ)/๐๐ฅ + ๐(2๐ 3๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ = 3. ๐2๐ฅ .๐(2๐ฅ)/๐๐ฅ + 2 .๐ 3๐ฅ . ๐(3๐ฅ)/๐๐ฅ ๐๐ฆ/๐๐ฅ = 3. ๐2๐ฅ . 2 + 2 .๐ 3๐ฅ. 3 ๐๐ฆ/๐๐ฅ = 6๐2๐ฅ + 6๐3๐ฅ ๐๐ฆ/๐๐ฅ = 6 (๐2๐ฅ + ๐3๐ฅ) Now, ๐ ๐/๐ ๐ = 6 (๐๐๐ + ๐๐๐) Again Differentiating ๐ค.๐.๐ก.๐ฅ (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 = (๐ (6(๐2๐ฅ" + " ๐3๐ฅ)))/๐๐ฅ (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 = 6 ๐(๐2๐ฅ" + " ๐3๐ฅ)/๐๐ฅ (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 = 6(๐(๐2๐ฅ)/๐๐ฅ + ๐(๐3๐ฅ)/๐๐ฅ) (๐^2 ๐ฆ)/ใ๐๐ฅใ^2 = 6(๐2๐ฅ. 2+๐3๐ฅ.3) (๐ ^๐ ๐)/ใ๐ ๐ใ^๐ = 6(๐๐๐๐+๐๐๐๐) Now we need to prove ๐ ๐๐/๐ ๐๐ โ 5 ๐ ๐/๐ ๐ + 6y = 0 Solving L.H.S ๐2๐ฆ/๐๐ฅ2 โ 5 ๐๐ฆ/๐๐ฅ + 6y = 6(2๐2๐ฅ+3๐3๐ฅ) โ 5.6 (๐2๐ฅ+๐3๐ฅ) + 6(3๐2๐ฅ+2๐3๐ฅ) = 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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Example 37 Important
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Example 40 You are here
Example 41
Example 42 Important Deleted for CBSE Board 2021 Exams only
Example 43 Deleted for CBSE Board 2021 Exams only
Example 44 Important
Example 45 Important
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Example 47 Important
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Chapter 5 Class 12 Continuity and Differentiability
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.