The solution of the differential equation `e^(-x) (y+1) dy +(cos^(2) x - sin 2x) y (dx) = 0` subjected to the condition that y = 1 when x = 0 is

Options

A. `y+logy+e^(x)cos^(2)x=2`
B. `log(y+1)+e^(x)cos^(2)x=1`
C. `y+logy=e^(x)cos^(2)x`
D. `(y+1)+e^(x)cos^(2)x=2`

Answer

a

Explanation

Given , ` e^(-x) (y+1) dy + (cos^(2) x - sin 2x ) ydx = 0 `
` rArr (1+1/y ) dy = - e^(x) (cos ^(2)x - sin 2 x) dx `
On integrating both sides , we get
` y + lo y = -e^(x) cos^(2) x + int e^(x) sin 2x dx`
` -int e^(x) sin 2x dx + C`
` rArr y + log y = -e^(x) cos^(2) x + C `
At x = 0 , y = 1
` 1+ 0 = - e^(0) cos 0 + C rArr C = 2`
` :. "Required solution is " y + log y = = -e^(x) cos^(2) x +2`