# Ex 9.6, 9

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 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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