# Ex 9.3, 5 - Chapter 9 Class 12 Differential Equations

Last updated at Dec. 10, 2019 by Teachoo

Last updated at Dec. 10, 2019 by Teachoo

Transcript

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))

Ex 9.3

Ex 9.3, 1
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Chapter 9 Class 12 Differential Equations

Serial order wise

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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.