# Ex 9.6, 7

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 9.6, 7 For each of the differential equation given in Exercises 1 to 12, find the general solution : 𝑥𝑙𝑜𝑔𝑥 𝑑𝑦𝑑𝑥+𝑦= 2𝑥𝑙𝑜𝑔𝑥 Step 1 : Put in form 𝑑𝑦𝑑𝑥 + Py = Q xlog x 𝑑𝑦𝑑𝑥 + y = 2𝑥 log x Dividing by x log x, 𝑑𝑦𝑑𝑥+𝑦. 1𝑥 log𝑥 = 2𝑥log 𝑥 1𝑥 log𝑥 𝑑𝑦𝑑𝑥 + 1𝑥 log𝑥𝑦= 2𝑥 Step 2 : Find P and Q Comparing (1) with 𝑑𝑦𝑑𝑥 + Py = Q P = 1𝑥 log𝑥 & Q = 2𝑥2 Step 3 : Find Integration factor, I.F I.F = e 𝑝 𝑑𝑥 I.F = e 1𝑥 log𝑥𝑑𝑥 Let t = log x dt = 1𝑥 dx dx = x dt So, I.F = e 1𝑥 𝑡 × 𝑥𝑑𝑡 I.F = e 1𝑡𝑑𝑡 I.F = e log𝑡 I.F = 𝑡 Putting back t = log x I.F = log x Step 4 : Solution of the equation y × I.F = 𝑄×𝐼.𝐹. 𝑑𝑥+𝐶 Putting values, y × log x = 2𝑥2 . log x. dx + C Let I = 2 log𝑥 𝑥−2𝑑𝑥 I = 2 log x. 𝑥−2𝑑𝑥− 1𝑥 𝑥−2𝑑𝑥𝑑𝑥 I = 2 log x . 𝑥−1(−1) − 1𝑥 . ( 𝑥−1)(−1).dx = 2 − log x. 1𝑥 + 1 𝑥2.𝑑𝑥 = 2 − 1𝑥 .log x + 𝑥−1(−1) = 2 −1𝑥 .log x − 1𝑥 = −2𝑥 (1 + log x) Putting value of I in (2) y log x = I + C y. log x = − 2𝑥 (1 + log x) + C y log 𝒙 = −𝟐𝒙 (1 + log 𝒙) + 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+ 𝑥2𝑑𝑦+2𝑥𝑦 𝑑𝑥= cot𝑥 𝑑𝑥 𝑥≠0 Step 1: Put in form 𝑑𝑦𝑑𝑥 + Py = Q (1 + x2)dy + 2xy.dx = cot x. dx Dividing both sides by (1 + x2) 𝑑𝑦𝑑𝑥 + 2𝑥𝑦(1 + 𝑥2) 𝑑𝑥𝑑𝑥 = cot(𝑥)(1 + 𝑥2). 𝑑𝑥𝑑𝑥 𝑑𝑦𝑑𝑥 + 2𝑥(1 + 𝑥2) y = cot𝑥(1 + 𝑥2) Step 2 : Find P and Q Comparing (1) with 𝑑𝑦𝑑𝑥 + Py = Q where P = 2𝑥(1 + 𝑥2) & Q = cot𝑥(1 + 𝑥2) Step 3 : Find Integrating factor, I.F I.F. = 𝑒 𝑝 𝑑𝑥 I.F. = e 2𝑥 1 + 𝑥2 𝑑𝑥 Let t = 1 + x2 dt = 2x dx I.F. = e 𝑑𝑡𝑡 = elog t = t = (1 + x2) So, I.F = 1 + x2 Step 4 : Solution of the equation y × I.F = 𝑄×𝐼.𝐹.𝑑𝑥+𝐶 Putting values, y.(1 + x2) = cot𝑥(1 + 𝑥2) × (1+𝑥2).dx + c y.(1 + x2) = cot𝑥 𝑑𝑥+𝑐 y (1 + x2) = log sin𝑥 + C Dividing by (1 + x2) y = (1 + x2)−1 log 𝐬𝐢𝐧𝒙+𝒄 𝟏+x2−𝟏 is the general solution of the given equation Ex 9.6, 9 For each of the differential equation find the general solution : 𝑥 𝑑𝑦𝑑𝑥+𝑦−𝑥+𝑥𝑦 cot𝑥=0 𝑥≠0 Step 1: Put in form 𝑑𝑦𝑑𝑥 + Py = Q x 𝑑𝑦𝑑𝑥 + y − x + xy cot x = 0 Dividing both sides by x 𝑑𝑦𝑑𝑥 + 𝑦𝑥 − 1 + y cot x = 0 𝑑𝑦𝑑𝑥 + y 1𝑥+ cot𝑥 − 1 = 0 𝑑𝑦𝑑𝑥 + 1𝑥+ cot𝑥 y = 1 Step 2: Find P and Q Comparing (1) with 𝑑𝑦𝑑𝑥 + Py = Q P = 1𝑥 + cot x & Q = 1 Step 3: Find integrating factor, I.F. I.F. = e 𝑝 𝑑𝑥 = e 1𝑥 + cot𝑥𝑑𝑥 = e e 1𝑥 𝑑𝑥 + cot𝑥 𝑑𝑥 = 𝑒 log𝑥 + log sin𝑥 = 𝑒 log𝑥 sin𝑥 = x. sin x So, I.F. = x.sin x Step 4 : Solution of the equation y × I.F. = Q×𝐼𝐹𝑑𝑥 + C y (x sin x) = 𝑥. 𝑠𝑖𝑛 𝑥𝑑𝑥 Let I = 𝑥. sin𝑥.𝑑𝑥 I = x sin𝑥 𝑑𝑥− 1. sin𝑥 𝑑𝑥𝑑𝑥 = x (− cos x) − 1.(− cos𝑥)𝑑𝑥 = − x. cos x + cos𝑥 𝑑𝑥 = − 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 = 𝑥 cos𝑥𝑥 sin𝑥 + sin𝑥𝑥 sin𝑥 + 𝑐𝑥 sin𝑥 y = −cot x + 𝟏𝒙 + 𝒄𝒙 𝒔𝒊𝒏𝒙 Which is the general solution of the given differential equation

Example 1
Important

Ex 9.1, 11 Important

Ex 9.1, 12 Important

Example 7 Important

Ex 9.3, 7 Important

Ex 9.3, 10 Important

Example 13 Important

Ex 9.4, 14 Important

Example 17 Important

Example 18 Important

Ex 9.5, 8 Important

Ex 9.5, 15 Important

Example 22 Important

Ex 9.6, 7 Important You are here

Ex 9.6, 13 Important

Ex 9.6, 14 Important

Example 25 Important

Example 27 Important

Example 28 Important

Misc 6 Important

Misc 11 Important

Misc 12 Important

Misc 13 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

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