Question 5 - Formation of Differntial equation when general solution given - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Formation of Differntial equation when general solution given
Formation of Differntial equation when general solution given
Last updated at April 16, 2024 by Teachoo
Question 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))