Answer:
Option D
Explanation:
We have, differential equations,
$\log(\frac{dy}{dx})=x\Rightarrow\frac{dy}{dx}=e^{x}$
$\Rightarrow$ $ dy=e^{x}dx$
integrating on both sides we get
$\int dy=\int e^{x}dx$
$\Rightarrow $ $ y= e^{x}+ C$ ........(i)
On putting x=0, y=1 is Eq .(i) , we get
1= e0+ C $\Rightarrow$ C=0
Now, the particular solution of the given differential is y= ex