# Thread: [SOLVED] Differentiation through Seperation

1. ## [SOLVED] Differentiation through Seperation

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

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

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

2. Hello, Iceman75x!

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

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

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

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

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

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

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

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