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