Antiderrivative of ${d^2 y \over dx^2} = 1-x^2$ 
At any point $(x,y)$ on a curve, ${d^2 y \over dx^2} = 1-x^2$, and an equation of the tangent line to the curve at the point $(1,1)$ is $y=2-x$. Find an equation of the curve.

This is what I've done
$${d^2 y \over dx^2} = 1-x^2
\\ \int dy' = \int (1-x^2)dx
\\y' = x- {x^3 \over 3} +C
\\ {dy\over dx}=x- {x^3 \over 3} +C
\\\int dy =\int (x- {x^3 \over 3} +C)dx
\\y= {x^2 \over 2}-{1\over 3} \cdot {x^4 \over 4}+c_1x + c_2
\\ y= {x^2 \over 2}-{x^4 \over 12}+c_1x +c_2$$
Do I now substitute (1,1) in this? I don't this is right. Someone help me. Thank you!

Hint:
Since $y=2-x$ is the tangent to the curve at $(1,1)$ we have $y'(1)=-1$, so
\begin{align*}
1-\frac13+c_1&=-1\\
\frac12-\frac1{12}+c_1+c_2&=1
\end{align*}
