


Ex 9.6
Ex 9.6, 2
Ex 9.6, 3 Important
Ex 9.6, 4
Ex 9.6, 5 Important
Ex 9.6, 6
Ex 9.6, 7 Important
Ex 9.6, 8 Important You are here
Ex 9.6, 9
Ex 9.6, 10 Deleted for CBSE Board 2022 Exams
Ex 9.6, 11 Deleted for CBSE Board 2022 Exams
Ex 9.6, 12 Important Deleted for CBSE Board 2022 Exams
Ex 9.6, 13
Ex 9.6, 14 Important
Ex 9.6, 15
Ex 9.6, 16 Important
Ex 9.6, 17 Important
Ex 9.6, 18 (MCQ)
Ex 9.6, 19 (MCQ) Important Deleted for CBSE Board 2022 Exams
Last updated at Dec. 11, 2019 by Teachoo
Ex 9.6, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : (1+π₯^2 )ππ¦+2π₯π¦ ππ₯=cotβ‘γπ₯ ππ₯(π₯β 0)γ Given equation (1 + x2)dy + 2xy dx = cot x dx Dividing both sides by dx (1 + x2)ππ¦/ππ₯ + 2xy ππ₯/ππ₯ = cot x ππ₯/ππ₯ (1 + x2)ππ¦/ππ₯ + 2xy = cot x Dividing both sides by (1 + x2) ππ¦/ππ₯ + 2π₯/((1 + π₯2)) y = cotβ‘π₯/((1 + π₯2)) Comparing (1) with ππ¦/ππ₯ + Py = Q where P = 2π₯/((1 + π₯^2)) & Q = cotβ‘π₯/((1 + π₯^2)) Finding Integrating factor, I.F I.F. = π^β«1βγπ ππ₯γ I.F. = e^β«1βγ2π₯/((1 + π₯^2 ) ) ππ₯ γ Let t = 1 + x2 dt = 2x dx β¦(1) I.F. = e^β«1βγππ‘/π‘ γ = elog t = t = (1 + x2) So, I.F = 1 + x2 Solution of the equation is y Γ I.F = β«1βγπΓπΌ.πΉ.ππ₯+πΆγ Putting values, y.(1 + x2) = β«1βcotβ‘π₯/((1 + π₯2)) Γ (1+π₯2).dx + c y.(1 + x2) = β«1βγcotβ‘π₯ ππ₯γ+πΆ y (1 + x2) = log |sinβ‘π₯ | + C Dividing by (1 + x2) y = (1 + x2)β1 log |π¬π’π§β‘π |+πͺ(π+"x2" )^(βπ) is the general solution of the given equation Note: This answer does not match with the answer of the book. If we have made any mistake, please comment