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Question

Find a particular solution of the differential equation , given that y = 0 when

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Solution

The given differential equation is,

dy dx +ycotx=4xcosecx

The given differential equation is of the form dy dx +Py=Q.

Here,

P=cotx Q=4xcosecx

Now, calculate the Integrating factor,

IF= e Pdx = e cotx dx = e log| sinx | =sinx

The solution of the differential equation is,

y( IF )= ( Q×IF )dx +c ysinx= ( 4xcosecx×sinx ) dx+c ysinx= ( 4x ) dx+c ysinx=2 x 2 +c (1)

Given that y=0 when x= π 2 .

Substitute the values in the equation (1),

0×sin π 2 =2 ( π 2 ) 2 +c c= π 2 2

Substitute the value of c in the equation (1).

ysinx=2 x 2 π 2 2

Thus, the solution of the differential equation is ysinx=2 x 2 π 2 2 .


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