Find f. f '' (x) = 4 + 6x + 36x2, f (0) = 2, f (1) = 10

It has been a while since I have done differnetial equations but I think here you only need to integrate twice.

4 + 6x + 36x^2 = 2 * (2+3x+18x^2)

Integrate 2+3x+18x^2 to get 2x+3/2x^2+18/3x^3 + C = 2x + 3/2x^2+6x^3 + C

Integrate again now this: 2x + 3/2x^2+6x^3 + C to get x^2 + 1/2x^3+3/2x^4 + Cx + D

Now multiply by two (as I have taken out 2 earlier)

2x^2+x^3+3x^4+2Cx+D

Now use the initial conditions.

f (0) = 2

D=2

f (1) = 10

2+1+3+2C+2=10

C=1

So the solution is: (ordered by power, as neat mathematicans would do)

f (x) = 3x^4 + x^3 + 2x^2 + 2x + 2