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?
 A: Since $y(1)=1$ then by your answer we get
$$\frac A3+\frac B2+c+d=1\iff d=1-\frac A3-\frac B2-c$$
and then the solution is
$$y=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac A3-\frac B2-c\\=\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
A: 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
$$\frac{Ax^3}{3} + \frac{Bx^2}{2} + cx + \underbrace{1-\left(\frac{A}{3} + \frac{B}{2}+c\right)}_{d} =\\ =  \frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+1$$
A: $$y'=\text{Ax}^2+\text{Bx}+c$$
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d \ \ \  (*) $$
Using the initial condition $y(1)=1$ we get:
$$y(1)=1 \Rightarrow \frac{A }{3}+\frac{B }{2}+c +d=1 \Rightarrow d=1-\frac{A}{3}-\frac{B}{2}-c$$
Replacing this at the relation $(*)$ we get:
$$y(x)=\frac{A x^3}{3}+\frac{B x^2}{2}+c x+1-\frac{A}{3}-\frac{B}{2}-c=\frac{A x^3-A}{3}+\frac{B x^2-B}{2} +cx-c+1 \\ \Rightarrow y(x)=\frac{1}{3}A(x^3-1)+\frac{1}{2}B(x^2-1) +c(x-1)+1 $$
A: 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.
A: $$\frac{A x^3}{3}+\frac{B x^2}{2}+c x+d$$ where $d$ is a cosntant to fix is equivalent to $$\frac{1}{3} A \left(x^3-1\right)+\frac{1}{2} B \left(x^2-1\right)+c (x-1)+d$$ where $d$ is a constant to fix.
And the second one is surely more handy to apply the initial condition.
