Solving Bernoulli's Equation: Step-by-Step Guide

[domesticbark]

Homework Statement

<br /> 2xy&#039;+y^3e^(-2x)=2xy<br />

Homework Equations

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

The Attempt at a Solution

<br /> dy/dx-y=-y^3e^(-2x)/2x<br />

<br /> P(x)=-1<br /> Q(x)=-e^(-2x)/2x<br />

<br /> n=3<br />
<br /> v=1/y^2<br />

<br /> dy/dx=dy/dv*dv/dx<br />

<br /> dy/dx=-1/2v^-(3/2)*dv/dx<br />

<br /> -1/2v^(-3/2)*dv/dx-v^(-1/2)=-v^(-3/2)e^(-2x)/2x<br />

<br /> dv/dx + 2v=e^(-2x)/x<br />

<br /> e^(\int P(x)\,dx)=e^(\int -1\,dx)=e^(-x)<br />

<br /> dv/dx*e^(-x)+2ve^(-x)=e^(-3x)/x<br />


This is supposed to look live reverse chain rule so I get (e^(-x)*v)&#039;=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.

 

[LCKurtz]

Homework Statement

<br /> 2xy&#039;+y^3e^(-2x)=2xy<br />

Homework Equations

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

The Attempt at a Solution

<br /> dy/dx-y=-y^3e^(-2x)/2x<br />

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