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

Answer

D

Explanation

`e^(-x)(y+1)dy+(cos^(2)x-sin 2x) y (dx)=0`
`therefore" "(1+y^(-1))dy + e^(x)(cos^(2)x-sin 2x)dx = 0`
Integration we get,
`y+log_(e) y + e^(x) cos^(2) x = c`
y = 1 when x = 0
`therefore" "1+0+1 = c`
`therefore" "c = 2`
`therefore" ""Curve is y" + log y + e^(x) cos^(2) x = 2`