Example 37 - Chapter 5 Class 12 Continuity and Differentiability
Last updated at Dec. 16, 2024 by Teachoo
Examples
Example 1
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Transcript
Example 37 If y = 3e2x + 2e3x, prove that 𝑑2𝑦/𝑑𝑥2 − 5 𝑑𝑦/𝑑𝑥 + 6y = 0. Given, 𝑦 = 3𝑒2𝑥 + 2𝑒3𝑥 Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(3𝑒2𝑥 + 2𝑒3𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 𝑑(3𝑒 2𝑥)/𝑑𝑥 + 𝑑(2𝑒 3𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 3. 𝑒2𝑥 .𝑑(2𝑥)/𝑑𝑥 + 2 .𝑒 3𝑥 . 𝑑(3𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥 = 3. 𝑒2𝑥 . 2 + 2 .𝑒 3𝑥. 3 𝑑𝑦/𝑑𝑥 = 6𝑒2𝑥 + 6𝑒3𝑥 𝑑𝑦/𝑑𝑥 = 6 (𝑒2𝑥 + 𝑒3𝑥) Now, 𝒅𝒚/𝒅𝒙 = 6 (𝒆𝟐𝒙 + 𝒆𝟑𝒙) Again Differentiating 𝑤.𝑟.𝑡.𝑥 (𝑑^2 𝑦)/〖𝑑𝑥〗^2 = (𝑑 (6(𝑒2𝑥" + " 𝑒3𝑥)))/𝑑𝑥 (𝑑^2 𝑦)/〖𝑑𝑥〗^2 = 6 𝑑(𝑒2𝑥" + " 𝑒3𝑥)/𝑑𝑥 (𝑑^2 𝑦)/〖𝑑𝑥〗^2 = 6(𝑑(𝑒2𝑥)/𝑑𝑥 + 𝑑(𝑒3𝑥)/𝑑𝑥) (𝑑^2 𝑦)/〖𝑑𝑥〗^2 = 6(𝑒2𝑥. 2+𝑒3𝑥.3) (𝒅^𝟐 𝒚)/〖𝒅𝒙〗^𝟐 = 6(𝟐𝒆𝟐𝒙+𝟑𝒆𝟑𝒙) Now we need to prove 𝒅𝟐𝒚/𝒅𝒙𝟐 − 5 𝒅𝒚/𝒅𝒙 + 6y = 0 Solving L.H.S 𝑑2𝑦/𝑑𝑥2 − 5 𝑑𝑦/𝑑𝑥 + 6y = 6(2𝑒2𝑥+3𝑒3𝑥) − 5.6 (𝑒2𝑥+𝑒3𝑥) + 6(3𝑐2𝑥+2𝑒3𝑥) = 12𝑒2𝑥 + 18𝑒3𝑥 − 30𝑒2𝑥 − 30𝑒3𝑥 + 18𝑒2𝑥 + 12𝑒3𝑥 = 12𝑒2𝑥 − 30𝑒2𝑥 + 18𝑒2𝑥 + 18𝑒3𝑥 − 30𝑒3𝑥 + 12𝑒3𝑥 = 30𝑒2𝑥 − 30𝑒2𝑥 + 30𝑒3𝑥 − 30𝑒3𝑥 = 0 =RHS Hence proved
