Question

If $\frac{d y}{d x}-y \log _{e} 2=2^{\sin x}(\cos x-1) \log _{e} 2,$ then $y=$

• A. $2^{\sin x}+c 2^{x}$
• B. $2^{\cos x}+c 2^{x}$
• C. $2^{\sin x}+c 2^{-x}$
• D. $2^{\cos x}+c 2^{-x}$

Answer 1

Answer: (A)

$\frac{d y}{d x}-y \log _{e} 2=2^{\sin x}(\cos x-1) \log _{e} 2$
This is linear differential equation
I.F. $=e^{-\log _{e} 2 \int d x}=e^{-x \log _{e} 2}=2^{-x}$
Solution is
$y 2^{-x}=\int 2^{-x} 2^{\sin x}(\cos x-1) \log _{e} 2\, d x$
put $\sin x-x=t \Rightarrow (\cos x-1) d x=d t$
$\therefore y 2^{-x}=\log _{e} 2 \int 2^{t} d t$
$ \therefore y 2^{-x}=2^{t}+c $
$\therefore y=2^{x+t}+c 2^{x} $
$\therefore y=2^{\sin x}+c 2^{x}$