What is the equation of the curve passing through the point ` (0,pi/3)` satisfying the differential equation ltbr? Sin x cos y dx + cos x sin y dy =0 ?

Options

A. `cos x cos y = sqrt3/2`
B. `sinx sin y = sqrt3/2`
C. `sin x siny =1/2`
D. ` cos x cos y = 1/2`

Answer

D

Explanation

Given differnetial equation is ltbr. Sin x cos y dx + cos x sin x dy =0
` Rightarrow sin x cos y dx = - cos x siny dy `
` Rightarrow (sinx )/(cos x) dx = - (siny)/(cos y) dy`
Integrate on both side
` int( sinx)/(cos x) dx = ( sin y)/(cos y) dy `
` Rightarrow -log (cosx ) =log ( cos y) + log c`
where log c is constant of integration
` Rightarrow -log c = log (cos y) + log ( cos x)`
` 1/c = cos y cos x`
Since, this curve passing through `( 0,pi/3)`
it satisfies equation (1)
So, ` 1/c = cos ""pi/3, cos 0`
` 1/c = 1/2 xx 1 Rightarrow c=2`
Hence, required equation of curve is cos x cos y =` 1/2`