The particular solution of the differential equation `log((dy)/(dx))=x`, then x=0, y=1 is . . .

Options

A. `y=e^(x)+2`
B. `y=-e^(x)`
C. `y=-e^(x)+2`
D. `y=e^(x)`

Answer

D

Explanation

We have, differential equations, ,brgt `log((dy)/(dx))=ximplies(dy)/(dx)=e^(x)`
`impliesdy=e^(x)dx`
On integrating both sides, we get
`intdy=inte^(x)dx`
`impliesy=e^(x)+C` . . . (i)
On putting x=0, y=1 is Eq. (i), we get
`1=e^(0)+C impliesC=0`
Now, particular solution of the given differential is `y=e^(x)`