Find general and singular solution of $px=3p^2e^y+1$, $p =\frac{dy}{dx}$
My attempt:-
$$px=(3p^2e^y+1),\qquad x= 3pe^y+\frac{1}{p}$$
differentiating w.r.to $y$
$$1/p = 3e^y \frac{dp}{dy}+3pe^y-\frac{1}{p^2}\frac{dp}{dy}
\\~\\\implies \frac{1-3p^2 e^y}{p}= \frac{3e^yp^2-1}{p^2} \frac{dp}{dy}
\\~\\\implies \frac{dp}{p}+dy=0, p = ce^{-y},$$ $c$ is arbitrary constant
putting this value of $p$ in given equation, i got $$3p^2e^y-px+1=0 \implies 3c^2e^{-y}-ce^{-y}x+1=0$$ is required general solution
But textbook exercise answer is $3c^2e^{3y}-xce^y+1=0$ which means $p=ce^y$
I got $p=ce^{-y}$. i verified but could not find mistake. Pls have a look at this.
Best Answer
Set $u=e^{y}$, then $u'=e^{y}p$ so that the equation translates to $$ xu' = 3u'^2+u\iff u = xu'-3u'^2 $$ which is a Clairaut equation with the solution family of lines $u'=c$, $$ u = cx-3c^2\implies y=\ln(cx-3c^2) $$ and the singular solution $$ 0=x-6u'\iff u'=\frac x6\implies u=\frac{x^2}{12},\quad y=\ln(x^2)-\ln(12). $$
This confirms your solution, the text book is wrong, perhaps working from an equation where some sign is flipped. One can not say more without seeing the solution steps of the text book solution.