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Ex 9.3
Ex 9.3, 2 Deleted for CBSE Board 2023 Exams
Ex 9.3, 3 Important Deleted for CBSE Board 2023 Exams
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Ex 9.3, 10 Important Deleted for CBSE Board 2023 Exams
Ex 9.3, 11 (MCQ) Deleted for CBSE Board 2023 Exams
Ex 9.3, 12 (MCQ) Important Deleted for CBSE Board 2023 Exams
Last updated at Dec. 10, 2019 by Teachoo
Maths Crash Course - Live lectures + all videos + Real time Doubt solving!
Ex 9.3, 5 Form a differential equation representing the given family of curves by eliminating arbitrary constants π and π. π¦=π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ ) Since it has two variables, we will differentiate twice π¦=π^π₯ (π cosβ‘γπ₯+π sinβ‘π₯ γ ) Differentiating Both Sides w.r.t. π₯ ππ¦/ππ₯=π/ππ₯ [π^π₯ (π cosβ‘π₯+π sinβ‘π₯ )] π¦^β²=π(π^π₯ )/ππ₯.[π cosβ‘π₯+π sinβ‘π₯]+π^π₯ π/ππ₯ [π cosβ‘π₯+π sinβ‘π₯] π¦^β²=π^π₯ [π cosβ‘π₯+π sinβ‘π₯]+π^π₯ [βπ sinβ‘π₯+π cosβ‘π₯] π¦^β²=π¦+π^π₯ [βπ sinβ‘π₯+π cosβ‘π₯] π¦^β²βπ¦=π^π₯ [βπ sinβ‘π₯+π cosβ‘π₯] β¦(1) Again Differentiating both sides w.r.t.x π¦^β²β²βπ¦^β²=π(π^π₯ )/ππ₯ [βπ sinβ‘π₯+π cosβ‘π₯]+π^π₯ π/ππ₯ [βπ sinβ‘π₯+π cosβ‘π₯] π¦^β²β²βπ¦^β²=π^π [βπ πππβ‘π+π πππβ‘π]+π^π₯ [βπ cosβ‘π₯+π (βsinβ‘π₯)] π¦^β²β²βπ¦^β²=γ(πγ^β²β π)+π^π₯ [βπ cosβ‘π₯βπ sinβ‘π₯] π¦^β²β²βπ¦^β²=π¦^β²βπ¦βπ^π [π πππβ‘π+π πππβ‘π] π¦^β²β²βπ¦^β²=π¦^β²βπ¦βπ¦ π¦^β²β²βπ¦^β²=π¦^β²β2π¦ π¦^β²β²βπ¦^β²βπ¦^β²+2π¦=0 π¦^β²β²β2π¦^β²+2π¦=0 which is the required differential equation (From (1)) (Using y = π^π₯ (π cos x + b sin x))