<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] Use the condition $y(1)=1$, i.e. substitute $x=1$ and then you have 
\begin{equation}
\frac{A}{3} + \frac{B}{2} + c +d = 1
\end{equation}
So, you get 
\begin{equation}
d = 1 -\frac{A}{3} - \frac{B}{2} - c
\end{equation}
By substituting in $\frac{A}{3}x^3+\frac{B}{2}x^2+cx+d$, you get that equation.
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