<s>[INST] Integrate $\int \left(A x^2+B x+c\right) \, dx$ I am asked to find the solution to the initial value problem:
$$y'=\text{Ax}^2+\text{Bx}+c,$$
where $y(1)=1$,
I get:
$$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$
But the answer to this is: 
$$y=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1.$$
Could someone show me what has been done and explain why?
 [/INST] Both solutions are correct provided your $d$ satisfies the condition $y(1) = 1$. Note that in your solution $y(1) = \frac{A}{3} + \frac{B}{2} + c + d = 1$ so setting $d = 1 - (\frac{A}{3} + \frac{B}{2} + c)$ would be the way to go.
Note also that
$$\frac{Ax^3}{3} + \frac{Bx^2}{2} + cx + \underbrace{1-\left(\frac{A}{3} + \frac{B}{2}+c\right)}_{d} =\\ =  \frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
</s>