If `(dy)/(dx)-y log_(e) 2 = 2^(sin x)(cos x -1) log_(e) 2`, then y =

Options

A. `2^(sin x) + c2^(x)`
B. `2^(cos x)+c2^(x)`
C. `2^(sin x) + c2^(-x)`
D. `2^(cos x) + c2^(-x)`

Answer

A

Explanation

`(dy)/(dx)-y log_(e) 2 = 2^(sin x) (cos x - 1) log_(e) 2`
This is linear differential equation
I.F. `=e^(-log_(e) 2int dx) = e^(-x log_(e) 2) = 2^(-x)`
Solution is
`y 2^(-x) = int 2^(-x) 2^(sin x) (cos x - 1) log_(e) 2 dx`
put `sin x - x = t rArr (cos x -1) dx = dt`
`therefore" "y 2^(-x) = log_(e) 2 int 2^(t) dt`
`therefore" "y 2^(-x) = 2^(t) + c`
`therefore" "y = 2^(x+t) + c 2^(x)`
`therefore" "y = 2^(sin x) + c2^(x)`