<s>[INST] 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!
 [/INST] Since the equation of the tangent line to the curve at $(1,1)$ is $y=-x+2$, we know that $dy/dx|_{(x,y)=(1,1)}=-1=1-\frac{1}{3}+C \Rightarrow C=-\frac{5}{3}$
Thus, $y=\frac{x^2}{2}-\frac{x^4}{12}-\frac{5}{3}x+D$ and we know that 
$1=\frac{1}{2}-\frac{1}{12}-\frac{5}{3}+D\Rightarrow D=-\frac{9}{4}$
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