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!
 A: 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 
$1=\frac{1}{2}-\frac{1}{12}-\frac{5}{3}+D\Rightarrow D=-\frac{9}{4}$
A: 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*}
A: Don't get overwhelmed by the problem having multiple pieces You've accounted for the most complicated part of the problem, and now your goal is to solve

Suppose a curve is defined by $y = \frac{1}{2}x^2 - \frac{1}{12} x^4 + c_1 x + c_2$ for unknown constants $c_1$ and $c_2$.
Suppose also that this curve has a tangent line at $(1,1)$ given by $y = 2 - x$.
Find $c_1$ and $c_2$.

This could very well have been a word problem in your introductory calculus course — your job now is to forget you're in a differential equations course and remember how you solved problems like this in calculus.
A: You're doing it right. 


*

*$C$ and $c_1$ are the same thing. Don't confuse yourself!

*Substitute $(1,1)$ into the equation for $y$ to deduce the value of $c_2$.

*The gradient at $(1,1)$ is $-1$. So substitute that into the equation for $\frac{dy}{dx}$ and you'll get the value of $C$.
