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? 
 A: At stationary point $(3,2)$, $\frac{dy}{dx} = 0$
$$ (3x^2-a)^2 =0 $$
$$(27 -a)^2 = 0$$
$$a = 27$$
Therefore
$$ \frac{dy}{dx} = (3x^2 - 27)^2 = 9x^4 -162x^2+729 $$
If you solve the above differential equation, you should be able to find the general equation required-use $(3, 2)$ to find the constant of integration
A: 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 $c=2-\frac{8181}{5}=\frac{-8171}{5}=-1634.2$
A: $$\frac{dy}{dx}=(3x^2-a)^2=9x^4-6x^2a+a^2$$
$$y=a^2 x-2 a x^3+(9 x^5)/5 + constant$$
Having found that $a=27$ since $dy/dx=0$ at $(3,2)$, you use that same point to get the constant of integration. It turns out to be: $$-\frac{5822}{5}=-1164.4$$
A: $$\frac{d}{dx}[y]= \left(3x^2-a\right)^2$$
$$dy= \left(3x^2-a\right)^2dx$$
$$\int dy= \int\left(3x^2-a\right)^2dx$$
$$\int dy= \int\left(9x^4-6ax^2+a^2\right)dx$$
$$\int dy= 9\int x^4dx-6a\int x^2dx+a^2\int dx$$
$$y= \frac95 x^5-2ax^3+a^2x+C$$
Try your problem again and update your post with your latest attempt.
