Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 9.5

Ex 9.5, 1
Important

Ex 9.5, 2

Ex 9.5, 3 Important

Ex 9.5, 4

Ex 9.5, 5 Important

Ex 9.5, 6

Ex 9.5, 7 Important

Ex 9.5, 8 Important

Ex 9.5, 9 You are here

Ex 9.5, 10

Ex 9.5, 11

Ex 9.5, 12 Important

Ex 9.5, 13

Ex 9.5, 14 Important

Ex 9.5, 15

Ex 9.5, 16 Important

Ex 9.5, 17 Important

Ex 9.5, 18 (MCQ)

Ex 9.5, 19 (MCQ) Important

Last updated at May 29, 2023 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 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