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Question

으4(sec x) y= tan x| 0ㄨㄑㄧ

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Solution

The given differential equation is dy dx +secx×y=tanx,( 0x< π 2 ).

The above equation is in the form of dy dx +Py=Q.

Here the value is P=secx,Q=tanx.

Integrating Factor is calculated as follows,

IF= e Pdx = e secxdx = e log( secx+tanx ) =secx+tanx

The solution of the given differential equation is,

y( IF )= ( Q×IF ) dx+C y( secx+tanx )= tanx( secx+tanx ) dx+C y( secx+tanx )= secxtanx dx+ tan 2 x dx+C =secx+ ( sec 2 x1 ) dx+C

Further simplify.

y( secx+tanx )=secx+tanxx+C

Therefore, the above equation is required solution of differential equation.


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