<s>[INST] Find equation of curve ${dy \over dx}= (3x^2-a)^2$, where $a$ is a constant. Given that the curve has a stationary point at $(3,2)$, find the equation of the curve. 
I managed to get the equation $y=3x^3+3ax^2+xa^2$+c. I'm not sure if I integrated correctly though. If I did how do I proceed? 
 [/INST] We see that $\frac{\partial y}{\partial x} = (3x^2-a)^2= 9 x^4 - 6x^2a+a^2$. Taking the anti-derivative(integration) we see that $y=\frac{36}{5}x^2 - 2x^3a+a^2x + c$ where c is a constant. We know that the stationary point is $(3,2)$ and the derivative equals zero there(definition stationary point). So $\frac{\partial y}{\partial x}(3,2)=(3\cdot 3^2-a)^2= 0$ We can easily see that $a=3\cdot 3^2=27$. Thus $y=\frac{36}{5}x^2 - 54x^3+729x + c$. Filling in the point $(3,2)$ we see $2=\frac{36}{5} 3^2 - 54 \cdot3^3+729 \cdot 3 + c=\frac{324}{5}- 486+ 2187+c=\frac{8181}{5}+c$. 
Thus $c=2-\frac{8181}{5}=\frac{-8171}{5}=-1634.2$
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