Find the particular solution of the differential equation `(dy)/(dx) = 1+x+y+xy,` given that y = 0 when x = 1.

Answer

`log|1+y|=x+(x^(2))/(2)-(3)/(2)`

Explanation

`(dy)/(dx)=(1+x)(1+y)rArr int (dy)/((1+y))=int (1+x)dx rArr log|1+y|=x+(x^(2))/(2)+C " " `...(i)
Putting x = 1 and y = 0 in (i), we get `C=(-3)/(2).`
`therefore log|1+y|=x=(x^(2))/(2)-(3)/(2).`