# 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 22 Find 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 ) = ^ ^ /

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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.