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

y''=A(Ais a constant)

Integrating again gives:

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

Once more:

`y = (Ax^2)/2 + Bx + C` (

A, BandCare 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 givesC= 3.

and

y''(2) = 6 givesA= 6(Actually,

y''= 6 for any value ofxin this problem since there is noxterm)

Finally,

y'(1) = 4 givesB= -2.

So the particular solution for this question is:

y =3x^{2}−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.

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