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
