Answer 1

`dy/dx=1+x+y+xy`

`dy/dx` = 1 + x + y + xy

`dy/dx = 1( 1 + x ) + y ( 1 + x)`

`dy/dx=(1+x)(1+y)`        

`dy/(1+y)=(1+x)dx`

Integrating both sides:

`intdy/(1+y)=int(1+x)dx`

`log|1+y|=x+x^2/2+C`

y = 0 when  x = 1     (given)

`log1=1+1/2+C`

`C=−3/2`

`⇒log|1+y|=x+x^2−3/2` is the required solution.