What is the equation of the curve passing through the origin and satisfying the differential equation
dy = (ytanx+secx)dx ?

Options

A. y=x cos x
B. y cos x=x
C. xy=cos x
D. y sin x=x

Answer

A

Explanation

The differnetial equation
`dy = y(tna x +sec x) ` dx can be written as
` (dy)/(dx) = y tan x =sec x`
`(dy)/(dx) = y tan x + sec c`
` or, (dy)/(dx) -y tan x = sec x`
which is of the from ` (dy)/(dx) + P (x) .y = Q (x)`
Here `P(x) = - tan x and Q (x) = sec x`
Integrating factor IF = ` eint - (sin x)/(cos x) dx `
Putting cos x= t
-s in x dx =dt
`if = eint dt/t = e^(log_(e)t) = t= cos x `
the solution is
` y. Q(x) = int I.F.Q (x ) dx +c`
or, ` y.sec x=int cos x . sec x dx +c`
` or, y.sec x = int dx +c`
or, y.sec x = ` int dx +c`
or, y.sec x =x +c
Since the curve passes through the origin.
` 0=0+c Rightarrow c =0`
`and y sec x =x `
or , y =x cos x `