Since y''' = 0, when we integrate once we get:

y'' = A (A is a constant)

Integrating again gives:

y' = Ax + B (A, B are constants)

Once more:

`y = (Ax^2)/2 + Bx + C` (A, B and C are constants)

The boundary conditions are:

y(0) = 3, y'(1) = 4, y''(2) = 6

We need to substitute these values into our expressions for y'' and y' and our general solution, `y = (Ax^2)/2 + Bx + C`.

Now

y(0) = 3 gives C = 3.

and

y''(2) = 6 gives A = 6

(Actually, y'' = 6 for any value of x in this problem since there is no x term)

Finally,

y'(1) = 4 gives B = -2.

So the particular solution for this question is:

y = 3x² − 2x + 3

Checking the solution by differentiating and substituting initial conditions:

y' = 6x − 2

y'(1) = 6(1) − 2 = 4

y'' = 6

y''' = 0

Our solution is correct.