# Ex 9.5, 9 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Ex 9.5

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Ex 9.5, 9 You are here

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Ex 9.5, 18 (MCQ)

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Last updated at April 16, 2024 by Teachoo

Ex 9.5, 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 π π/π π + (π/π+πππβ‘π ) y = 1 Comparing (1) with ππ¦/ππ₯ + Py = Q P = π/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 Solution of the equation is y Γ I.F. = β«1βγQΓπΌπΉγβ‘ππ₯ + C y (x sin x) = β«1βγπ.γπππ πγβ‘π π γ Let I = β«1βγπ.π¬π’π§β‘γπ.π πγ 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 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 πππβ‘π)/(π πππβ‘π ) + πππβ‘π/(π πππβ‘π ) + πͺ/(π πππβ‘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 = 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 πππβ‘π)/(π πππβ‘π ) + πππβ‘π/(π πππβ‘π ) + πͺ/(π πππβ‘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