Bernoulli's Equation

  • #1

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.
 

Answers and Replies

  • #2
LCKurtz
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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]

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

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