Example 22 - Particular solution dy/dx + y cot x = 2x + x2 cot x - Examples

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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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