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Question 1 Deleted for CBSE Board 2025 Exams
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 3 Deleted for CBSE Board 2025 Exams
Question 4 Important Deleted for CBSE Board 2025 Exams
Question 5 Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Last updated at April 16, 2024 by Teachoo
Example 36 If π¦ = A sinβ‘π₯+B cosβ‘π₯, then prove that π2π¦/ππ₯2 + y = 0.π¦ = A sinβ‘π₯+B cosβ‘π₯ Differentiating π€.π.π‘.π₯ ππ¦/ππ₯ = π(A sinβ‘π₯ + B cosβ‘π₯" " )/ππ₯ ππ¦/ππ₯ = π(A sinβ‘π₯ )/ππ₯ + π(B cosβ‘π₯ )/ππ₯ ππ¦/ππ₯ = A . π(sinβ‘π₯ )/ππ₯ + B . π(cosβ‘π₯" " )/ππ₯ ππ¦/ππ₯ = A cosβ‘π₯" " + B (β sinβ‘π₯) π π/π π = A πππβ‘π" " β B πππβ‘π Again Differentiating π€.π.π‘.π₯ (π^2 π¦)/γππ₯γ^2 = (π (γA cosγβ‘π₯" " " β" γB sinγβ‘π₯))/ππ₯ (π^2 π¦)/γππ₯γ^2 = π(A cosβ‘π₯ )/ππ₯ β π(B sinβ‘π₯" " )/ππ₯ (π^2 π¦)/γππ₯γ^2 = βA sinβ‘π₯ β B cosβ‘π₯ (π^2 π¦)/γππ₯γ^2 = β (A sinβ‘π₯ + B cosβ‘π₯) (π^2 π¦)/γππ₯γ^2 = βy π ππ/π ππ + π = π Hence proved (As π¦ = π΄ π ππβ‘π₯+π΅ πππ β‘π₯)