Answer:
Option D
Explanation:
We have, differential equations,
log(dydx)=x⇒dydx=ex
⇒ dy=exdx
integrating on both sides we get
∫dy=∫exdx
⇒ y=ex+C ........(i)
On putting x=0, y=1 is Eq .(i) , we get
1= e0+ C ⇒ C=0
Now, the particular solution of the given differential is y= ex