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Example 40 - Chapter 5 Class 12 Continuity and Differentiability (Term 1)
Last updated at April 13, 2021 by Teachoo
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Example 40 You are here
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Transcript
Example 40 If y = 3e2x + 2e3x, prove that π2π¦/ππ₯2 β 5 ππ¦/ππ₯ + 6y = 0. Given, π¦ = 3π2π₯ + 2π3π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(3π2π₯ + 2π3π₯)/ππ₯ ππ¦/ππ₯ = π(3π 2π₯)/ππ₯ + π(2π 3π₯)/ππ₯ ππ¦/ππ₯ = 3. π2π₯ .π(2π₯)/ππ₯ + 2 .π 3π₯ . π(3π₯)/ππ₯ ππ¦/ππ₯ = 3. π2π₯ . 2 + 2 .π 3π₯. 3 ππ¦/ππ₯ = 6π2π₯ + 6π3π₯ ππ¦/ππ₯ = 6 (π2π₯ + π3π₯) Now, π π/π π = 6 (πππ + πππ) Again Differentiating π€.π.π‘.π₯ (π^2 π¦)/γππ₯γ^2 = (π (6(π2π₯" + " π3π₯)))/ππ₯ (π^2 π¦)/γππ₯γ^2 = 6 π(π2π₯" + " π3π₯)/ππ₯ (π^2 π¦)/γππ₯γ^2 = 6(π(π2π₯)/ππ₯ + π(π3π₯)/ππ₯) (π^2 π¦)/γππ₯γ^2 = 6(π2π₯. 2+π3π₯.3) (π ^π π)/γπ πγ^π = 6(ππππ+ππππ) Now we need to prove π ππ/π ππ β 5 π π/π π + 6y = 0 Solving L.H.S π2π¦/ππ₯2 β 5 ππ¦/ππ₯ + 6y = 6(2π2π₯+3π3π₯) β 5.6 (π2π₯+π3π₯) + 6(3π2π₯+2π3π₯) = 12π2π₯ + 18π3π₯ β 30π2π₯ β 30π3π₯ + 18π2π₯ + 12π3π₯ = 12π2π₯ β 30π2π₯ + 18π2π₯ + 18π3π₯ β 30π3π₯ + 12π3π₯ = 30π2π₯ β 30π2π₯ + 30π3π₯ β 30π3π₯ = 0 =RHS Hence proved
