# Math Help - [SOLVED] Differentiation through Seperation

1. ## [SOLVED] Differentiation through Seperation

Trying to solve $y'=(e^X) (Y^3)$

I will post all my work so hopefully I can figure out what I'm doing wrong.

$dy/dx=e^xy^3$
$y^-3 dy$=∫ $e^X dx$
$y^-2/2=e^x+C1$
$2/y^2=e^x+C$
$1/y^2=(e^x+C)/2$
$y^2=2/(e^x+C)$
$Y=sqrt(2/(e^x+C)$

2. Hello, Iceman75x!

Trying to solve: . $\frac{dy}{dx}\:=\:e^xy^3$

I will post all my work so hopefully I can figure out what I'm doing wrong.

$\frac{dy}{dx}\:=\:e^xy^3$

$y^{-3}\,dy\:=\:e^x\,dx$

$\int y^{-3}dy \:=\:\int e^x\,dx$

$\frac{y^{-2}}{2} \:=\:e^x + C_1$ . . . . here ... and the next step, too

It should have been: . $\frac{y^{-2}}{{\color{red}-}2} \:=\:e^x + C_1 \quad\Rightarrow\quad y^{-2} \:=\:-2e^x + C$

. . $\frac{1}{y^2} \:=\:C - 2e^x\quad\Rightarrow\quad y^2 \:=\:\frac{1}{C-2e^x} \quad\Rightarrow\quad\boxed{ y \;=\;\pm\frac{1}{\sqrt{C-2e^x}}}$