Question

Solve the differential equation:

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

Answer 1

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

=(1 + x)+ y (1+x)

= (1 + x) (1 + y)

∴ `dy/(1+y) = (1+x) dx`

Integrating on both sides, we get

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

∴ `log | 1+y| = x + x^2/2 + c`