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

Options

A. ` cos y = (cos a)/(sqrt(2))`
B. ` cos y = (sin x)/2`
C. ` cos y = (sec x)/(sqrt(2))`
D. ` cos y = ("cosec"x)/(sqrt(2))`

Answer

c

Explanation

The differential equation of the given curve is
`sin x cos y dx +cos x sin y dy =0`
`rArr (sinx)/(cosx) dx +(siny)/(cosy) dy = 0`
`rArr tan x dx + tan y dy = 0 `
On integrating both sides , we get
` int tan x dx _ int tan y dy = log C`
` rArr log ( sec x) + log (sec y ) = log C`
` sec x sec y = C " " ` ...(i)
The curve passes through the point `(0,pi/4)` ,therefore put
` x = 0 , y = pi/4,` we get ` sec 0 sec pi/4 = C`
` rArr C = sqrt(2)`
On putting the value of C in Eq. (i) , we get
` sec x * sec y = sqrt(2)`
` rarr sec x* 1/(cosy) = sqrt(2)`
` rArr cos y = (sec x)/(sqrt(2))`
Hence, the required equation of the curve is
` cos y =(secx)/(sqrt(2))`