1.

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`.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» Correct Answer - c
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))`


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