# Bernoulli's Equation

domesticbark

## Homework Statement

$2xy'+y^3e^(-2x)=2xy$

## Homework Equations

$dy/dx + P(x)y=Q(x)y^n$
$v=y^(1-n)$

## The Attempt at a Solution

$dy/dx-y=-y^3e^(-2x)/2x$

$P(x)=-1 Q(x)=-e^(-2x)/2x$

$n=3$
$v=1/y^2$

$dy/dx=dy/dv*dv/dx$

$dy/dx=-1/2v^-(3/2)*dv/dx$

$-1/2v^(-3/2)*dv/dx-v^(-1/2)=-v^(-3/2)e^(-2x)/2x$

$dv/dx + 2v=e^(-2x)/x$

$e^(\int P(x)\,dx)=e^(\int -1\,dx)=e^(-x)$

$dv/dx*e^(-x)+2ve^(-x)=e^(-3x)/x$

This is supposed to look live reverse chain rule so I get $(e^(-x)*v)'=e^(-3x)/x$
but it doesn't look right and I have no idea how to do the integral required to then solve the rest of this problem.

Homework Helper
Gold Member

## Homework Statement

$2xy'+y^3e^(-2x)=2xy$

## Homework Equations

$dy/dx + P(x)y=Q(x)y^n$
$v=y^(1-n)$

## The Attempt at a Solution

$dy/dx-y=-y^3e^(-2x)/2x$

That should be ##\frac{dy}{dx}-y = -y^3e^{-2x}##