# Ex 9.6, 5 - Chapter 9 Class 12 Differential Equations (Term 2)

Last updated at Aug. 20, 2021 by Teachoo

Ex 9.6

Ex 9.6, 1
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Ex 9.6, 5 Important You are here

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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, 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 Aug. 20, 2021 by Teachoo

Ex 9.6, 5 For each of the differential equation given in Exercises 1 to 12, find the general solution : cos2𝑥 𝑑𝑦𝑑𝑥+𝑦=𝑡𝑎𝑛𝑥 0≤𝑥< 𝜋2 Step 1: Put in form 𝑑𝑦𝑑𝑥 + Py = Q cos2x. 𝑑𝑦𝑑𝑥 + y = tan x Dividing by cos2x, 𝑑𝑦𝑑𝑥 + y. 1𝑐𝑜𝑠2𝑥 = tan𝑥𝑐𝑜𝑠2𝑥 ⇒ 𝑑𝑦𝑑𝑥 + (sec2x)y = sec2x. tan x Step 2: Find P and Q Comparing (1) with 𝑑𝑦𝑑𝑥 + Py = Q P = sec2 x and Q = sec2 x. tan x Step 3 : Find integrating factor, I.F I.F = e 𝑝𝑑𝑥 I.F = e 𝑠𝑒𝑐2𝑥.𝑑𝑥 I.F. = etan x Step 4 : Solution of the equation y × I.F. = 𝑄×𝐼.𝐹.𝑑𝑥+𝑐 Putting values, y.etan x = 𝑠𝑒𝑐2𝑥. tan𝑥.etan x.dx + C Let I = 𝑠𝑒𝑐2𝑥. tan𝑥.etan x.dx Putting t = tan x ⇒ 𝑠𝑒𝑐2𝑥.dx = dt Putting values of t & dt in equation ∴ I = tan𝑥.etan x.(𝑠𝑒𝑐2𝑥.𝑑𝑥) I = 𝑡. 𝑒𝑡.𝑑𝑡 I = t 𝑒𝑡𝑑𝑡 − 𝑑𝑡𝑑𝑡 𝑒𝑡𝑑𝑡𝑑𝑡 I = t.et − et 𝑑𝑡 I = 𝑡et − et . Putting t = tan x I = tan x. etan x – etan x I = etan x ( tan x − 1) Substituting value of I in (2), y etan x = etan x (tan x − 1) + C Dividing by etan x, y = tan x − 1 + C. e–tan x