Find the equation of the curve passing through the point `(0,pi/4)` whose differential equation is `s in" "x" "cos" "y" "dx" "+" "cos" "x" "s in" "y" "dy" "=" "0` .

Explanation

Differential equation of given curve is :
`sin xcosydx+cosxsinydy=0`
`implies(sinx)/(cosx)dx+(siny)/(cosy)dy=0`
`implies tanxdx+tanydy=0`
On integration , `inttanxdx+inttanydy=logC`
`implieslog(secx)+log(secy)=logC`
`implies secx*secy=C`…….`(1)`
Curve passes through the point `(0,(pi)/(4))`, so put `x=0`, `y=(pi)/(4)`
`sec0sec"(pi)/(4)=CimpliesC=sqrt(2)`
put the value of `C` in equation `(1)` ,
`secx*secy=sqrt(2)`
`implies secx*(1)/(cosy)=sqrt(2)implies cosy=(secx)/(sqrt(2))`
Therefore, required equation of curve is : `cosy=(secx)/(sqrt(2))`.