# Ex 9.6, 8 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Dec. 11, 2019 by Teachoo

Ex 9.6

Ex 9.6, 1
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Ex 9.6, 2

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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

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Ex 9.6, 14 Important

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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

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