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Last updated at Dec. 11, 2019 by Teachoo

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Misc 18 The general solution of the differential equation π^π₯ ππ¦+(π¦ π^π₯+2π₯)ππ₯=0 is (A) π₯ π^π¦+π₯^2=πΆ (B) π₯ π^π¦+π¦^2=πΆ (C) π¦ π^π₯+π₯^2=πΆ (D) π¦ π^π¦+π₯^2=πΆ Given equation π^π₯ ππ¦+(π¦ π^π₯+2π₯)ππ₯=0 π^π₯ ππ¦=β(π¦ π^π₯+2π₯)ππ₯ ππ¦/ππ₯= (β(π¦π^π₯ + 2π₯))/π^π₯ ππ¦/ππ₯ = (βπ¦π^π₯)/π^π₯ β2π₯/π^π₯ ππ¦/ππ₯ = βπ¦β2π₯/π^π₯ ππ¦/ππ₯ + y = (β2π₯)/π^π₯ Differential equation is of the form ππ¦/ππ₯ + Py = Q where P = 1 & Q = (β2π₯)/π^π₯ IF = π^β«1βγπ ππ₯γ IF = π^β«1βγ1 ππ₯γ IF = π^π₯ Solution is y(IF) = β«1βγ(πΓπΌπΉ)ππ₯+πγ yex = β«1βγ(β2π₯)/π^π₯ π^π₯ ππ₯+πγ yex = ββ«1βγ2π₯ ππ₯+πγ yex = β2Γπ₯^2/2+π yex = βπ₯^2+π yex + π₯^2=π β΄ Option (C) is the correct answer

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 9 years. He provides courses for Maths and Science at Teachoo.