Question

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

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