# Ex 9.6, 7 - Chapter 9 Class 12 Differential Equations

Last updated at Nov. 14, 2019 by Teachoo

Last updated at Nov. 14, 2019 by Teachoo

Transcript

Ex 9.6, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : + = 2 Step 1 : Put in form + Py = Q xlog x + y = 2 log x Dividing by x log x, + . 1 log = 2 log 1 log + 1 log = 2 Step 2 : Find P and Q Comparing (1) with + Py = Q P = 1 log & Q = 2 2 Step 3 : Find Integration factor, I.F I.F = e I.F = e 1 log Let t = log x dt = 1 dx dx = x dt So, I.F = e 1 I.F = e 1 I.F = e log I.F = Putting back t = log x I.F = log x Step 4 : Solution of the equation y I.F = . . + Putting values, y log x = 2 2 . log x. dx + C Let I = 2 log 2 I = 2 log x. 2 1 2 I = 2 log x . 1 ( 1) 1 . ( 1 ) ( 1) .dx = 2 log x. 1 + 1 2 . = 2 1 .log x + 1 ( 1) = 2 1 .log x 1 = 2 (1 + log x) Putting value of I in (2) y log x = I + C y. log x = 2 (1 + log x) + C y log = (1 + log ) + C Which is the general solution of the given equation. Ex 9.6, 8 For each of the differential equation given in Exercises 1 to 12, find the general solution : 1+ 2 +2 = cot 0 Step 1: Put in form + Py = Q (1 + x2)dy + 2xy.dx = cot x. dx Dividing both sides by (1 + x2) + 2 (1 + 2) = cot ( ) (1 + 2 ) . + 2 (1 + 2) y = cot (1 + 2) Step 2 : Find P and Q Comparing (1) with + Py = Q where P = 2 (1 + 2 ) & Q = cot (1 + 2 ) Step 3 : Find Integrating factor, I.F I.F. = I.F. = e 2 1 + 2 Let t = 1 + x2 dt = 2x dx I.F. = e = elog t = t = (1 + x2) So, I.F = 1 + x2 Step 4 : Solution of the equation y I.F = . . + Putting values, y.(1 + x2) = cot (1 + 2) (1+ 2).dx + c y.(1 + x2) = cot + y (1 + x2) = log sin + C Dividing by (1 + x2) y = (1 + x2) 1 log + +x2 is the general solution of the given equation Ex 9.6, 9 For each of the differential equation find the general solution : + + cot =0 0 Step 1: Put in form + Py = Q x + y x + xy cot x = 0 Dividing both sides by x + 1 + y cot x = 0 + y 1 + cot 1 = 0 + 1 + cot y = 1 Step 2: Find P and Q Comparing (1) with + Py = Q P = 1 + cot x & Q = 1 Step 3: Find integrating factor, I.F. I.F. = e = e 1 + cot = e e 1 + cot = log + log sin = log sin = x. sin x So, I.F. = x.sin x Step 4 : Solution of the equation y I.F. = Q + C y (x sin x) = . Let I = . sin . I = x sin 1. sin = x ( cos x) 1.( cos ) = x. cos x + cos = x cos x + sin x Putting value of I in (2), y.x. sin x = x.cos x + sin x + C Divide by x sin x y = cos sin + sin sin + sin y = cot x + + Which is the general solution of the given differential equation

Chapter 9 Class 12 Differential Equations

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.