The general solution of differential equation `(dy)/(dx)=e^((x^(2))/(2))+xy` is

Options

A. `(y+1)=k(e^(x)+1)`
B. `y+1=e^(x)+1+k`
C. `y=log {(y+1)(e^(x)+1)}`
D. `y=log{(e^(x)+1)/(y+1)}+k`

Explanation

Given differential equation
`" "(e^(x)+1)ydy=(y+1)e^(x)dx`
`rArr" '(dy)/(dx)=(e^(x)(1+y))/((e^(x)+)y)rArr(dx)/(dy)=((e^(x)+1)y)/(e^(x)(1+y))`
`rArr" "(dx)/(dy)=(e^(x)y)/(e^(x)(1+y))+(y)/(e^(x)(1+y))`
`rArr" "(dx)/(dy)=(y)/(1+y)+(y)/((1+y)e^(x))`
`rArr" "(dx)/(dy)=(y)/(1+y)(1+(1)/(e^(x)))`
`rArr" "(dx)/(dy)=(y)/(1+y)((e^(x)+1)/(e^(x)))`
`rArr" "((y)/(1+y))dy=((e^(x))/(e^(x)+1))dx`
On integrating both sides, we get
`" "int(y)/(1+y)dy=int(e^(x))/(1+e^(x))dx` `rArr" "int(1+y-1)/(1+y)dy=int(e^(x))/(1+e^(x))dx`
`rArr" "int1dy-int(1)/(1+y)dy=int(e^(x))/(1+e^(x))dx`
`rArr" "y-log|(1+y)=log|(1+e^(x))+logk`
`rArr" "y=log(1+y)+log(1+e^(x))+log(k)`
`rArr" "y=log{k(1+y)(1+e^(x))}`