# Ex 9.6, 9 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Ex 9.6

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Ex 9.6, 9 You are here

Ex 9.6, 10 Deleted for CBSE Board 2022 Exams

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Ex 9.6, 12 Important Deleted for CBSE Board 2022 Exams

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Ex 9.6, 18 (MCQ)

Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams

Chapter 9 Class 12 Differential Equations (Term 2)

Serial order wise

Last updated at Dec. 11, 2019 by Teachoo

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Ex 9.6, 9 For each of the differential equation find the general solution : π₯ ππ¦/ππ₯+π¦βπ₯+π₯π¦ cotβ‘γπ₯=0(π₯β 0)γ Given equation 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 β¦(1) Comparing (1) with ππ¦/ππ₯ + Py = Q P = 1/π₯ + cot x & Q = 1 Finding integrating factor, I.F. I.F. = e^β«1βγπ ππ₯ γ = e^β«1β(1/π₯ + cotβ‘π₯ )ππ₯ = e^β«1βγ1/π₯ ππ₯ + β«1βγcotβ‘π₯ ππ₯γγ = π^(logβ‘π₯ + logβ‘sinβ‘π₯ ) = π^logβ‘γ(π₯ sinβ‘π₯)γ = x sin x So, I.F. = x sin x (Using log a + log b = log ab) (Using π^logβ‘π₯ = x) Solution of the equation is 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¦(2) Using formula β«1βγπ(π₯)π(π₯)ππ₯=π(π₯)ππ(π₯)ππ₯ββ«1β[πβ²(π₯)][π(π₯)ππ₯] γ dx Taking f(x) = x & g(x) = sin x = 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 = (βπ₯ cosβ‘π₯)/(π₯ sinβ‘π₯ ) + sinβ‘π₯/(π₯ sinβ‘π₯ ) + πΆ/(π₯ sinβ‘π₯ ) y = βcot x + 1/π₯ + πΆ/(π₯ π ππβ‘π₯ ) y = π/π β cot x + πͺ/(π πππβ‘π ) Which is the general solution of the given differential equation