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Example 40 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
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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Chapter 5 Class 12 Continuity and Differentiability (Term 1)
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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.