- #1

domesticbark

- 6

- 0

## Homework Statement

[itex]

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

[/itex]

## Homework Equations

[itex]

dy/dx + P(x)y=Q(x)y^n

[/itex]

[itex]

v=y^(1-n)

[/itex]

## The Attempt at a Solution

[itex]

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

[/itex]

[itex]

P(x)=-1

Q(x)=-e^(-2x)/2x

[/itex]

[itex]

n=3

[/itex]

[itex]

v=1/y^2

[/itex]

[itex]

dy/dx=dy/dv*dv/dx

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

This is supposed to look live reverse chain rule so I get [itex] (e^(-x)*v)'=e^(-3x)/x [/itex]

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.