# Example 22 - Chapter 9 Class 12 Differential Equations

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Example 22Find the particular solution of the differential equation / + cot =2 + ^2 cot ( 0) given that =0 = /2 / + cot =2 + ^2 cot Differential equation is of the form / + = where P = cot x & Q = 2x + x2 cot x IF = ^ 1 IF = ^ 1 cot IF = ^"log sin x " IF = sin x Solution is y (IF) = 1 ( ) + y sin x 1 sin =(2 + ^(2 ) cot ) + y sin x = 1 (2 sin + ^(2 ) sin cot ) + y sin = 1 2 sin + 1 ^2 sin cot + y sin = 2 1 sin ( ) + ^2 sin cot + y sin = 2 [sin 1 1 [cos 1 ] ] + 1 ^2 sin cos /sin dx + C y sin x = 2 sin x ["sin x " [ ^2/2]" 2" 1 cos x [ ^2/2] + 1 ^2 cos " dx + c" ] y sin x = x2sin x 1 ^2 cos x dx + 1 ^2 cos x dx + c y sin x = x2 sin x + c Given that y = 0 when x = /2 Putting = /2 and y = 0 in (1) (0) sin /2=( /2)^2 sin ( /2)+C 0 = ^2/4 (1)+C ^2/4=C Putting value in C in (1) y sin x = ^2 sin ^2/4 Dividing whole by sin x ( sin )/sin =( ^2 sin )/sin ^2/(4 sin ) = ^ ^ /

Example 1
Important

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7 Important

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13 Important

Example 14

Example 15

Example 16

Example 17 Important

Example 18 Important

Example 19

Example 20

Example 21

Example 22 Important You are here

Example 23

Example 24

Example 25 Important

Example 26

Example 27 Important

Example 28 Important

Chapter 9 Class 12 Differential Equations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.