<s>[INST] Calculus I: Tangent line at certain point, second derivative of the curve given. How to solve? I have never seen this problem in my homework.. I need help right before final!!
Let $\frac{d^2y}{dx^2}=y''=-x^3$ at every point on a curve. The equation of the tangent line at $(1,1)$ is $y = 3-2x$. Find the equation of the curve.
If someone can help me out with this problem, will greatly appreciate. Thank you.
 [/INST] Observe that via integration, it follows that 
$$
y' = - \frac{x^4}{4} + C.
$$
Hence, at $(1,1)$, $y'(1) = C - \frac{1}{4}$, and the tangent line is of the form $y - 1 = (  C - \frac{1}{4} )(x - 1)$, that is ( using the knowledge of the tangent curve at (1,1) ) that
$$
 3 - 2x = y = \left(C- \frac{1}{4} \right)x + \frac{5}{4} - C. 
$$ 
Then, $C - \frac{1}{4} = - 2 \implies C = - \frac{7}{4}$. Also, we know that $\frac{5}{4} - C = 3 \implies C = - \frac{7}{4}$. Since these conditions agree, we have the solutions that 
$$
y' = - \frac{1}{4} \left(x^4 + 7 \right)
$$
so that $y$ is any curve of the form $y = - \frac{1}{20} x^5 - \frac{7}{4} x + K$, for real numbers $K$. Further, we know that $(1,1)$ is a point on the curve, and hence that $K$ must be such that $1 = - \frac{1}{20} - \frac{7}{4} + K$. It follows that $K = \frac{56}{20} = \frac{14}{5}$. So,
$$
y = - \frac{1}{20} x^5 - \frac{7}{4} x + \frac{14}{5} .
$$
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