# Ex 9.6, 9 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 9.6, 9For each of the differential equation find the general solution : π₯ ππ¦/ππ₯+π¦βπ₯+π₯π¦ cotβ‘γπ₯=0(π₯β 0)γ Step 1: Put in form ππ¦/ππ₯ + Py = Q x ππ¦/ππ₯ + y β x + xy cot x = 0 Dividing both sides by x ππ¦/ππ₯ + π¦/π₯ β 1 + y cot x = 0 ππ¦/ππ₯ + y (1/π₯+cotβ‘π₯ ) β 1 = 0 ππ¦/ππ₯ + (1/π₯+cotβ‘π₯ ) y = 1 Step 2: Find P and Q Comparing (1) with ππ¦/ππ₯ + Py = Q P = 1/π₯ + cot x & Q = 1 Step 3: Find integrating factor, I.F. I.F. = e^β«1βγπ ππ₯ γ = e^β«1β(1/π₯ + cotβ‘π₯ )ππ₯ = e e^β«1βγ1/π₯ ππ₯ + β«1βγcotβ‘π₯ ππ₯γγ = π^(logβ‘π₯ + logβ‘sinβ‘π₯ ) = π^logβ‘γπ₯ sinβ‘π₯ γ = x. sin x So, I.F. = x.sin x Step 4 : Solution of the equation y Γ I.F. = β«1βγQΓπΌπΉγβ‘ππ₯ + C y (x sin x) = β«1βγπ₯.γπ ππ π₯γβ‘ππ₯ γ Let I = β«1βγπ₯.sinβ‘γπ₯.ππ₯γ γ I = x β«1βsinβ‘γπ₯ ππ₯ββ«1β[1.β«1βsinβ‘γπ₯ ππ₯γ ]ππ₯γ = x (β cos x) β β«1βγ1.(βcosβ‘γπ₯)γ ππ₯γ = β x. cos x + β«1βcosβ‘γπ₯ ππ₯γ = β x cos x + sin x Putting value of I in (2), y.x. sin x = βx.cos x + sin x + C Divide by x sin x y = (π₯ cosβ‘π₯)/(π₯ sinβ‘π₯ ) + sinβ‘π₯/(π₯ sinβ‘π₯ ) + π/(π₯ sinβ‘π₯ ) y = βcot x + π/π + π/(π πππβ‘π ) Which is the general solution of the given differential equation

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.