Answer 1

We have,
y' − 2x − 2 = 0 .....(1)
Now,
y = x² + 2x + C
⇒ y' = 2x + 2
Putting the above value in (1), we get
LHS = 2x + 2 − 2x − 2 = 0 = RHS
Thus, y = x² + 2x + C is the solution of the given differential equation.