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Findthe 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`. |
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Answer» 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))`. |
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