<s>[INST] How to solve differential equation $3p^2e^y-px+1=0$ ,$p =\frac{dy}{dx}$ How to solve differential equation
$$3p^2e^y-px+1=0$$
where $$p =\frac{dy}{dx}$$
I have tried to solve for p and for x, but i am not getting anywhere. Can someone help me with this
Thanks
 [/INST] By using the quadratic formula we get
\begin{align*}
3p^2e^y-px+1&=0\\
3p^2-pe^{-y}x+e^{-y}&=0\\
p&=\frac{xe^{-y}\pm\sqrt{x^2e^{-2y}-12e^{-y}}}{6}\\
pe^y&=\frac{x\pm\sqrt{x^2-12e^{y}}}{6}\\
6pe^y-x&=\pm\sqrt{x^2-12e^{y}}
\end{align*}
Let $u=-6e^y+\frac{1}{2}x^2$, then we have
\begin{align*}
-u'&=\pm\sqrt{2u}\\
u^{-1/2}u'&=\mp\sqrt{2}\\
\end{align*}
Integrating respect to $x$ 
$$2u^{1/2}=\mp\sqrt{2}x+c\implies 4u=2(x+k)^2$$
Here $c,\;k$ are constants, then
$$-24e^y+2x^2=2(x+k)^2$$ 
$$-24e^y=4k x+2k^2$$
Finally we get
$$\boxed{\color{blue}{y=\log\left(-\frac{1}{6}kx-\frac{1}{12}k^2\right)}}$$
</s>